Coordinate geometry

1. Cartesian plane

Exercises

2. Straight line graphs

Calculating a line from two points

You are given the following diagram:

Calculate the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the gradient m of the line and then the y-intercept (c). Recall that the the equation of a straight line is of the form y=mx+c.


STEP: Calculate the gradient m of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(4)(3,5)(2)(3)m=1,5

STEP: Calculate the y-intercept (c) of the line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituing a point A into the general form for a straight line:

y=mx+c(3,5)=(1,5)×(3)+cc=1

STEP: Substitute the values of m and c into the equation y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=1,5x+1

Submit your answer as:

Determining the equation of a straight line

In the diagram below, the line passes through the point A(1;0,5) and has a gradient m=1,5.

Find the equation of the line.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of the equation of a straight line:

y=mx+c

In the question statement we are given m, so the only unknown is c.

y=(1,5)x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

We need to calculate the value of the y-intercept (c) of the line. We do this by substituing the coordinates of the given point into the general form for a straight line:

y=(1,5)x+c0,5=(1,5)(1)+cc=2

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the straight line is:

y=1,5x2

Submit your answer as:

The gradient-intercept form of the straight line equation

Given the following diagram of a straight line passing through the points A(2;1,5) and B(1;0).

Determine the equation of the line AB.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient m of the line:

m=yByAxBxA=(0)(1,5)(1)(2)=0,5

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c). We do this by substituing the coordinates of point A into the general form for a straight line:

y=mx+c1,5=(0,5)×(2)+cc=0,5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is

y=0,5x+0,5

Submit your answer as:

Finding the equations of parallel lines

Consider the diagram given below:

Line AB is parallel to y=0,5x2 (the green line).

Determine the equation of AB.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of AB, we first need to determine the gradient of the line. We know that if two lines are parallel, then they must have equal gradients:

m1=m2=0,5General equation:y=mx+cy=0,5x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we calculate the value of the c by substituting the coordinates of point A into the equation:

y=0,5x+c3,5=(0,5)(3)+cc=2

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is

y=0,5x+2

Submit your answer as:

Calculating a gradient from two points

You are given the diagram below. The coordinates are A:(2;2,5) and B:(2;0,5).

Calculate the gradient (m) of line AB, correct to 2 decimal places.

Answer: Gradient (m) = .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Calculate the gradient m using the formula:

m=yByAxBxA,

where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the gradient formula
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the gradient of the line shown. We can use the gradient formula to do this.

The gradient formula uses the coordinates of two points (remember that the symbol for gradient is m):

m=y2y1x2x1

STEP: Substitute the coordinates of points A and B into the gradient formula
[−1 point ⇒ 0 / 2 points left]

Now we substitute the coordinates of the points into the formula to calculate the gradient. Remember to use brackets around the numbers when you substitute to keep all of the negative signs organised.

m=(0,5)(2,5)(2)(2)m=24m=0,5

Therefore the gradient m of the line AB is 0,5.


Submit your answer as:

Finding an unknown coordinate on a line

Consider the diagram below, which shows a straight line. The line passes through the points A(8;52) and B(x;4).

The equation of the line is: y=x2+32.

Determine the x-coordinate of Point B.

INSTRUCTION: Your answer should be exact (do not round off).
Answer: The x-coordinate of Point B is: .
numeric
STEP: Use the equation of the line
[−2 points ⇒ 0 / 2 points left]

We have a line passing through two points, but Point B has an unknown coordinate. We can use the equation of the line to find the value of the missing coordinate.

Remember that every point on the line solves the equation (the coordinates of the point make the equation true). So substitute B(x;4) - the coordinates of B - into the equation and solve for the unknown value. To make it a bit more clear, we can use xB to show that the coordinate we want belongs to Point B.

y=x2+32(4)=xB2+3252=xB25=xB
NOTE: We did not need the coordinates of Point A for any part of this solution. Since we already know the equation of the line, we can use Point B immediately in the equation.

The x-coordinate of Point B is 5.


Submit your answer as:

Finding the coordinate of a point using parallel lines

The following diagram shows two parallel lines. Point A is at (5;3) and Point B is at (1;y). Line AB runs parallel to the following line: y=x+3, which is the dashed blue line.

Determine the y-coordinate of Point B.

Answer: Point B = (1;).
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the fact that parallel lines have equal gradients.


STEP: Start building the equation for line AB
[−1 point ⇒ 2 / 3 points left]

We need to find the y-coordinate of Point B. We do not know much about line AB. But we do know that it is parallel to the line y=x+3. That means the lines have the same gradient! We can read the gradient from the equation.

y=x+3gradient=1

Line AB has the same gradient, so we can start building the equation for the line:

y=mx+cyAB=1x+c=x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

We know the coordinates of Point A. We can use them with this equation to calculate the value of c.

y=x+c(3)=1(5)+cc=2

Great! Now we have the equation for line AB:yAB=x2.


STEP: Substitute the coordinates of Point B into the equation and solve for yB
[−1 point ⇒ 0 / 3 points left]

Finally, we substitute the coordinates of Point B into the equation for line AB:

y=x2(yB)=(1)2

Solving, we get:

yB=3
NOTE:

You could also solve this question using the gradient formula. The gradient of line AB is −1 so we could set up this equation using the coordinates from Points A and B:

m=y2y1x2x11=yB31(5)

If you solve this equation, you will get the same answer that we got using the equation of line AB.

The y-coordinate of Point B is 3.


Submit your answer as:

Working with parallel lines

The diagram below shows two parallel lines. The dashed red line has the equation:

ydashed=x4+112

The solid blue line passes through the points A(x;54) and B(8;52), as shown. Note that the x-coordinate of point A is unknown.

Determine xA, the x-coordinate of point A.

Answer: xA=
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the fact that the two lines must have equal gradients because they are parallel. So you can find the gradient of line AB by looking at the equation for the other line.


STEP: Use the fact that parallel lines have equal gradients
[−1 point ⇒ 3 / 4 points left]

The two lines in the graph are parallel. That means they must have equal gradients. The equation for the red dashed line is:

y=x4+112

which means that both of the lines have a gradient of 14. So the equation of line AB must be:

y=x4+c

STEP: Find the value of c for the line using point B
[−1 point ⇒ 2 / 4 points left]

In this question we need a missing coordinate from point A. However, we know both coordinates for point B, so we can use them to find the value of c in the equation for line AB.

y=x4+c(52)=(8)4+cof point B to find cUse the coordinates12=c

Now we have the complete equation for line AB:

y=x4+12

STEP: Use the y-coordinate of point A to find xA
[−2 points ⇒ 0 / 4 points left]

Now we can substitute the y-coordinate for point A into the equation for line AB and solve for xA. This works because the equation of any line agrees with all of the points on the line: if we substitute in the y-coordinate then when we solve for x we will get the coordinate for the point which belongs on the line: the equation forces this to be true!

yA=xA4+12(54)=xA4+12xA=7

The correct answer is xA=7.


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Finding the equation of a straight line

The diagram below shows a line. The line has a gradient m=32 and it passes through the point A(1;72).

Find the equation of the line. Give only the right side of the equation, following "y= " below.

Answer: Equation of the line: y =
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You should start with the standard form for a straight line (y=mx+c) and substitute in the gradient. Then use the coordinates of point A to find c.


STEP: Start building the equation using the gradient
[−1 point ⇒ 1 / 2 points left]

The first thing to remember is that a straight line has this general form:

y=mx+c

where m is the gradient and c is the y-intercept. We have the gradient m of the line (it is given in the question) so we can substitute that into the equation of the line:

y=(32)x+cy=3x2+c

STEP: Use the coordinates of the point to find c
[−1 point ⇒ 0 / 2 points left]

Now we need the value of c for the line. We can find this by substituing the coordinates of any point into the general form for a straight line. This works because every point on the line solves the equation (the coordinates of the points agree with the equation). Since we have the coordinates of point A we will use them.

y=3x2+c(72)=3(1)2+c72=32+c2=c

Notice that c is positive: this agrees with the diagram, because the y-intercept is above the x-axis.

The equation of the line is:

y=3x2+2

Submit your answer as:

Finding a point on a line

You are given the following diagram:

Point C lies on the line AB.

Calculate the missing coordinate of Point C.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point C = (;2,1).
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the equation of the straight line AB.


STEP: Calculate the gradient (m) of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the unknown variable on the line, we start by calculating the equation of the straight line. We first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(1,5)(3,5)(3)(2)m=1

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting Point A into the general form for a straight line:

y=mx+c(3,5)=(1)×(2)+cc=1,5

STEP: Solve for x using the known value of Point C
[−1 point ⇒ 0 / 3 points left]

We now substitute the known values of Point C into the known equation of the straight line:

y=mx+cy=(1)x+1,52,1=(1)(x)+1,5x=0,6

Submit your answer as:

Finding a coordinate of a point using an equation

You are given the following diagram. It shows point A at (−2;1,5) and point B at (1;y).

The equation of line segment AB is: y=x0,5

Determine the y coordinate of point B, correct to 2 decimal places.

Answer: Point B = (1;)
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the equation together with the coordinates of point B.


STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 2 / 3 points left]

We have two points on a line, but one of the coordinates is missing. We need to find that coordinate. The equation given in the question is the key to finding the answer.

y=x−0,5

Every point on the line must agree with this equation. That includes the coordinates of point B. We substitute the known value for point B into the equation because that will give us the value of the missing coordinate y which agrees with the equation.

y=x0,5y=(1)0,5

STEP: Solve the equation for the value of y
[−2 points ⇒ 0 / 3 points left]

Solving, we get:

y=1,5

Submit your answer as:

Finding the equation of a parallel line

Find the equation of a line passing through the point Z(3;3) which is parallel to y=3x54.

INSTRUCTION: Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.
Answer:
equation
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. You can take the gradient of the first line straight from the equation: y=3x54. The gradient for this line is m=35.

For a line which is parallel to this one, the gradient of the line we want is also m=35.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now use the point (3;3), given together with the gradient, to get the equation of the line. Substitute the x and y-values into the equation to get the value of c:

y=mx+c3=(35)(3)+c3=95+c65=c

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is y=3x565.

The graph below shows the two equations: the blue line is the equation given in the question, and the red line is the answer. The point Z(3;3) is shown with a black dot and you can see that the red line passes through the point. You can also see that the two lines are parallel.


Submit your answer as:

Finding the equation of a straight line

The diagram below shows a straight line. The line passes through the coordinates A(2;6) and B(2;6). Calculate the equation of the line AB.

INSTRUCTION: Type the right side of the equation in the form mx+c.
Answer: y =
expression
STEP: Find the gradient of the line
[−1 point ⇒ 1 / 2 points left]

To find the equation of any straight line, we need the gradient of the line. We have two points on the line, so we can use the gradient formula.

m=yByAxBxA=(6)(6)(2)(2)=3

So the gradient of the line is 3.


STEP: Find the y-intercept of the line
[−1 point ⇒ 0 / 2 points left]

Now we need to find the y-intercept of the line. We do this by substituting the gradient and any point on the line into the general form for a straight line. Here we will use the coordinates of point A. (You can use point B if you want: any point on the line will work.)

y=mx+c(6)=(3)(2)+c6=6+c0=c

Therefore, the equation of the line AB is as follows:

y=3x

Submit your answer as:

Calculating the coordinates of a point on a line

Points A(1;1) and B(5;y) lie on the straight line y=0,5x+1,5.

Determine the y-coordinate of point B, rounded to 2 decimal places.

Answer: Point B = (5;)
numeric
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

We are given the equation of the straight line:

y=0,5x+1,5

To determine the value of y, we substitute the x-coordinate for point B into the equation:

y=0,5x+1,5y=(0,5)(5)+1,5

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We rearrange the equation to get the answer:

y=4

Submit your answer as:

Identifying parallel lines

Are the following lines parallel?

3y1=2x+116y=24x+2y+3

Note: In tests and exams you must show calculations to support your answer.

Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, you need to know the gradient of each equation. That means that must start by arranging the equations to be in standard form: y=mx+c. If the equations do not contain the variable y, then you need to isolate x to get an equation which says, "x= a number."
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

You can only answer this question by comparing the gradients of the two equations, which means you must arrange the equations to be in standard form. For the first equation this is:

3y1=2x+113y=2x+1213(3y)=13(2x+12)y=2x3+4

The gradient of the first equation is m=23.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now change the second equation to standard form.

6y=24x+2y+34y=24x+314(4y)=14(24x+3)y=6x+34

The gradient of the second equation is m=6.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Compare the two gradients to decide if the lines are parallel or not. For the equations in this question, the gradients are not equal, so the lines cannot be parallel. The correct response is: not parallel.

The two equations are graphed on the Cartesian plane shown below. You can see that the lines are not parallel.


Submit your answer as:

Calculating a coordinate for a point using the gradient

Consider the following diagram. Point A is at (−1;3,5) and Point B is at (2;y). Line AB has a gradient (m) of −1,5.

Calculate the y-coordinate of Point B.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point B= (2;)
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the equation of a straight line, which includes the gradient. Substitute in values that you know.


STEP: Find the equation of the line segment
[−2 points ⇒ 2 / 4 points left]

We have two endpoints of a line segment, and the gradient of the segment. However, the y-coordinate of Point B is missing. We need to find that coordinate value.

We can solve this question using the equation for a straight line:

y=mx+c

Firstly, we can calculate the value of the y-intercept (c) of the line AB. We do this by substituting the gradient value and the coordinates of Point A into the general form for a straight line:

y=mx+c(3,5)=(1,5)×(1)+cc=2

This means that the equation for the line segment in the graph is:

y=1,5x+2

STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known value for Point B into the equation:

y=1,5x+2y=1,5(2)+2

STEP: Solve the equation for y
[−1 point ⇒ 0 / 4 points left]

Solving, we get:

y=1

Submit your answer as:

Finding an equation with parallel lines

You are given the following diagram:

You are also told that line segment AB runs parallel to the following line: y=1x3. Point A is at (3;4). Work out the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by recalling that the gradient of parallel lines is equal.


STEP: Determine the gradient of line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of the line AB. We can get the gradient from the parallel line. The parallel line has (by definition) the same gradient as the line AB:

y=mx+cy=(1)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(4)=(1)×(3)+cc=1

STEP: Write down the equation of the line AB in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=1x+1

Submit your answer as:

Find the equation of a straight line

The points Q and R have the coordinates (52;5) and (4;173), respectively. Find the equation of the straight line through Q and R. Your answer should be exact (no rounding).

INSTRUCTION: Type the entire equation into the answer box.
Answer: The equation is: .
one-of
type(equation)
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Think about the standard form of a straight line equation: is there anything we can substitute into the formula from the given information?
STEP: Substitute the values into the gradient formula
[−2 points ⇒ 3 / 5 points left]

The standard formula of a straight line is:

y=mx+c

where m is the gradient of the graph and c is the y-intercept. We are given only two points on the straight line. So we should use those points to try and find the slope and y-intercept of the equation. Gradient is basically determined by the ratio of vertical change over horizontal change, 'rise-over-run'. We are not given the graph, but we can always use the gradient formula:

m=y2y1x2x1

In this specific case we can represent each coordinate as follows:

x1=52x2=4y1=5y2=173

Now substitute these values into the formula to find the gradient.

gradient=(173)5(4)(52)=(23)(132)=(23)×(213)=(439)

STEP: Now determine the c-value
[−2 points ⇒ 1 / 5 points left]

If we look at the standard form of a straight line equation again, we now have:

y=(439)x+c
Where (439) is the gradient of the graph. Also notice that this value is negative. This means that the straight line is decreasing, moving from the top downwards. The next step is to solve for the c-value. In other words we need to find where the graph 'cuts' the y-axis. We can use any point on the straight line to substitute into our formula. Here we will use the given point: Q(52;5).
y=(439)x+c(5)=(439)(52)+c(5)+(1039)=cc=20539

The graph 'cuts' the y-axis at y=20539.


STEP: Now substitute all the calculated values into the standard formula
[−1 point ⇒ 0 / 5 points left]

The only thing left to do is to give the full equation of the straight line. We have solved all the missing parts that make up the formula of a straight line.

Therefore the equation is

y=439x+20539

Submit your answer as:

Exercises

Calculating a line from two points

You are given the following diagram:

Calculate the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the gradient m of the line and then the y-intercept (c). Recall that the the equation of a straight line is of the form y=mx+c.


STEP: Calculate the gradient m of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(4)(0,5)(5)(4)m=0,5

STEP: Calculate the y-intercept (c) of the line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituing a point A into the general form for a straight line:

y=mx+c(0,5)=(0,5)×(4)+cc=1,5

STEP: Substitute the values of m and c into the equation y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=0,5x1,5

Submit your answer as:

Calculating a line from two points

You are given the following diagram:

Calculate the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the gradient m of the line and then the y-intercept (c). Recall that the the equation of a straight line is of the form y=mx+c.


STEP: Calculate the gradient m of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(2)(4)(1)(3)m=1,5

STEP: Calculate the y-intercept (c) of the line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituing a point A into the general form for a straight line:

y=mx+c(4)=(1,5)×(3)+cc=0,5

STEP: Substitute the values of m and c into the equation y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=1,5x0,5

Submit your answer as:

Calculating a line from two points

You are given the following diagram:

Calculate the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the gradient m of the line and then the y-intercept (c). Recall that the the equation of a straight line is of the form y=mx+c.


STEP: Calculate the gradient m of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(4)(2)(2)(4)m=1

STEP: Calculate the y-intercept (c) of the line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituing a point A into the general form for a straight line:

y=mx+c(2)=(1)×(4)+cc=2

STEP: Substitute the values of m and c into the equation y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=1x+2

Submit your answer as:

Determining the equation of a straight line

In the diagram below, the line passes through the point A(2;3) and has a gradient m=0,5.

Calculate the equation of the line.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of the equation of a straight line:

y=mx+c

In the question statement we are given m, so the only unknown is c.

y=(0,5)x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

We need to calculate the value of the y-intercept (c) of the line. We do this by substituing the coordinates of the given point into the general form for a straight line:

y=(0,5)x+c3=(0,5)(2)+cc=2

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the straight line is:

y=0,5x+2

Submit your answer as:

Determining the equation of a straight line

In the diagram below, the line passes through the point A(2;2,5) and has a gradient m=1.

Determine the equation of the line.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of the equation of a straight line:

y=mx+c

In the question statement we are given m, so the only unknown is c.

y=(1)x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

We need to calculate the value of the y-intercept (c) of the line. We do this by substituing the coordinates of the given point into the general form for a straight line:

y=(1)x+c2,5=(1)(2)+cc=0,5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the straight line is:

y=x+0,5

Submit your answer as:

Determining the equation of a straight line

In the diagram below, the line passes through the point A(2;2) and has a gradient m=0,5.

Find the equation of the line.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of the equation of a straight line:

y=mx+c

In the question statement we are given m, so the only unknown is c.

y=(0,5)x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

We need to calculate the value of the y-intercept (c) of the line. We do this by substituing the coordinates of the given point into the general form for a straight line:

y=(0,5)x+c2=(0,5)(2)+cc=1

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the straight line is:

y=0,5x+1

Submit your answer as:

The gradient-intercept form of the straight line equation

Given the following diagram of a straight line passing through the points A(2;2,5) and B(3;2,5).

Calculate the equation of the line AB.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient m of the line:

m=yByAxBxA=(2,5)(2,5)(3)(2)=1

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c). We do this by substituing the coordinates of point A into the general form for a straight line:

y=mx+c2,5=(1)×(2)+cc=0,5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is

y=x+0,5

Submit your answer as:

The gradient-intercept form of the straight line equation

Given the following diagram of a straight line passing through the points A(2;2,5) and B(5;1).

Determine the equation of the line AB.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient m of the line:

m=yByAxBxA=(1)(2,5)(5)(2)=0,5

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c). We do this by substituing the coordinates of point A into the general form for a straight line:

y=mx+c2,5=(0,5)×(2)+cc=1,5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is

y=0,5x1,5

Submit your answer as:

The gradient-intercept form of the straight line equation

Given the following diagram of a straight line passing through the points A(2;2) and B(1;4).

Find the equation of the line AB.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first calculate the gradient m of the line:

m=yByAxBxA=(4)(2)(1)(2)=2

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c). We do this by substituing the coordinates of point A into the general form for a straight line:

y=mx+c2=(2)×(2)+cc=2

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is

y=2x2

Submit your answer as:

Finding the equations of parallel lines

Consider the diagram given below:

Line AB is parallel to y=0,5x1 (the green line).

Find the equation of AB.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of AB, we first need to determine the gradient of the line. We know that if two lines are parallel, then they must have equal gradients:

m1=m2=0,5General equation:y=mx+cy=0,5x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we calculate the value of the c by substituting the coordinates of point A into the equation:

y=0,5x+c1,5=(0,5)(1)+cc=2

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is

y=0,5x+2

Submit your answer as:

Finding the equations of parallel lines

Consider the diagram given below:

Line AB is parallel to y=1,5x5,5 (the green line).

Determine the equation of AB.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of AB, we first need to determine the gradient of the line. We know that if two lines are parallel, then they must have equal gradients:

m1=m2=1,5General equation:y=mx+cy=1,5x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we calculate the value of the c by substituting the coordinates of point A into the equation:

y=1,5x+c3=(1,5)(1)+cc=1,5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is

y=1,5x1,5

Submit your answer as:

Finding the equations of parallel lines

Consider the diagram given below:

Line AB is parallel to y=1,5x+4,5 (the green line).

Find the equation of AB.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of AB, we first need to determine the gradient of the line. We know that if two lines are parallel, then they must have equal gradients:

m1=m2=1,5General equation:y=mx+cy=1,5x+c

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we calculate the value of the c by substituting the coordinates of point A into the equation:

y=1,5x+c0=(1,5)(1)+cc=1,5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is

y=1,5x+1,5

Submit your answer as:

Calculating a gradient from two points

You are given the diagram below. The coordinates are A:(2;0) and B:(5;3,5).

Calculate the gradient (m) of line AB, correct to 2 decimal places.

Answer: Gradient (m) = .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Calculate the gradient m using the formula:

m=yByAxBxA,

where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the gradient formula
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the gradient of the line shown. We can use the gradient formula to do this.

The gradient formula uses the coordinates of two points (remember that the symbol for gradient is m):

m=y2y1x2x1

STEP: Substitute the coordinates of points A and B into the gradient formula
[−1 point ⇒ 0 / 2 points left]

Now we substitute the coordinates of the points into the formula to calculate the gradient. Remember to use brackets around the numbers when you substitute to keep all of the negative signs organised.

m=(3,5)(0)(5)(2)m=3,57m=0,5

Therefore the gradient m of the line AB is 0,5.


Submit your answer as:

Calculating a gradient from two points

You are given the diagram below. The coordinates are A:(4;3,5) and B:(1;1,5).

Calculate the gradient (m) of line AB, correct to 2 decimal places.

Answer: Gradient (m) = .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Calculate the gradient m using the formula:

m=yByAxBxA,

where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the gradient formula
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the gradient of the line shown. We can use the gradient formula to do this.

The gradient formula uses the coordinates of two points (remember that the symbol for gradient is m):

m=y2y1x2x1

STEP: Substitute the coordinates of points A and B into the gradient formula
[−1 point ⇒ 0 / 2 points left]

Now we substitute the coordinates of the points into the formula to calculate the gradient. Remember to use brackets around the numbers when you substitute to keep all of the negative signs organised.

m=(1,5)(3,5)(1)(4)m=55m=1

Therefore the gradient m of the line AB is 1.


Submit your answer as:

Calculating a gradient from two points

You are given the diagram below. The coordinates are A:(1;3,5) and B:(1;0,5).

Calculate the gradient (m) of line AB, correct to 2 decimal places.

Answer: Gradient (m) = .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Calculate the gradient m using the formula:

m=yByAxBxA,

where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the gradient formula
[−1 point ⇒ 1 / 2 points left]

The question asks us to find the gradient of the line shown. We can use the gradient formula to do this.

The gradient formula uses the coordinates of two points (remember that the symbol for gradient is m):

m=y2y1x2x1

STEP: Substitute the coordinates of points A and B into the gradient formula
[−1 point ⇒ 0 / 2 points left]

Now we substitute the coordinates of the points into the formula to calculate the gradient. Remember to use brackets around the numbers when you substitute to keep all of the negative signs organised.

m=(0,5)(3,5)(1)(1)m=32m=1,5

Therefore the gradient m of the line AB is 1,5.


Submit your answer as:

Finding an unknown coordinate on a line

Consider the diagram below, which shows a straight line. The line passes through the points A(8;2) and B(1;y).

The equation of the line is: y=x4.

Calculate the y-coordinate of Point B.

INSTRUCTION: Your answer should be exact (do not round off).
Answer: The y-coordinate of Point B is: .
numeric
STEP: Use the equation of the line
[−2 points ⇒ 0 / 2 points left]

We have a line passing through two points, but Point B has an unknown coordinate. We can use the equation of the line to find the value of the missing coordinate.

Remember that every point on the line solves the equation (the coordinates of the point make the equation true). So substitute B(1;y) - the coordinates of B - into the equation and solve for the unknown value. To make it a bit more clear, we can use yB to show that the coordinate we want belongs to Point B.

y=x4yB=(1)4yB=14
NOTE: We did not need the coordinates of Point A for any part of this solution. Since we already know the equation of the line, we can use Point B immediately in the equation.

The y-coordinate of Point B is 14.


Submit your answer as:

Finding an unknown coordinate on a line

Consider the diagram below, which shows a straight line. The line passes through the points A(1;83) and B(5;y).

The equation of the line is: y=x3+3.

Calculate the y-coordinate of Point B.

INSTRUCTION: Your answer should be exact (do not round off).
Answer: The y-coordinate of Point B is: .
numeric
STEP: Use the equation of the line
[−2 points ⇒ 0 / 2 points left]

We have a line passing through two points, but Point B has an unknown coordinate. We can use the equation of the line to find the value of the missing coordinate.

Remember that every point on the line solves the equation (the coordinates of the point make the equation true). So substitute B(5;y) - the coordinates of B - into the equation and solve for the unknown value. To make it a bit more clear, we can use yB to show that the coordinate we want belongs to Point B.

y=x3+3yB=(5)3+3yB=143
NOTE: We did not need the coordinates of Point A for any part of this solution. Since we already know the equation of the line, we can use Point B immediately in the equation.

The y-coordinate of Point B is 143.


Submit your answer as:

Finding an unknown coordinate on a line

Consider the diagram below, which shows a straight line. The line passes through the points A(6;12) and B(x;3).

The equation of the line is: y=x41.

Find the x-coordinate of Point B.

INSTRUCTION: Your answer should be exact (do not round off).
Answer: The x-coordinate of Point B is: .
numeric
STEP: Use the equation of the line
[−2 points ⇒ 0 / 2 points left]

We have a line passing through two points, but Point B has an unknown coordinate. We can use the equation of the line to find the value of the missing coordinate.

Remember that every point on the line solves the equation (the coordinates of the point make the equation true). So substitute B(x;3) - the coordinates of B - into the equation and solve for the unknown value. To make it a bit more clear, we can use xB to show that the coordinate we want belongs to Point B.

y=x41(3)=xB412=xB48=xB
NOTE: We did not need the coordinates of Point A for any part of this solution. Since we already know the equation of the line, we can use Point B immediately in the equation.

The x-coordinate of Point B is 8.


Submit your answer as:

Finding the coordinate of a point using parallel lines

The following diagram shows two parallel lines. Point A is at (1;1) and Point B is at (x;3). Line AB runs parallel to the following line: y=x5, which is the dashed blue line.

Determine the x-coordinate of Point B.

Answer: Point B = (;-3).
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the fact that parallel lines have equal gradients.


STEP: Start building the equation for line AB
[−1 point ⇒ 2 / 3 points left]

We need to find the x-coordinate of Point B. We do not know much about line AB. But we do know that it is parallel to the line y=x5. That means the lines have the same gradient! We can read the gradient from the equation.

y=x5gradient=1

Line AB has the same gradient, so we can start building the equation for the line:

y=mx+cyAB=1x+c=x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

We know the coordinates of Point A. We can use them with this equation to calculate the value of c.

y=x+c(1)=1(1)+cc=2

Great! Now we have the equation for line AB:yAB=x2.


STEP: Substitute the coordinates of Point B into the equation and solve for xB
[−1 point ⇒ 0 / 3 points left]

Finally, we substitute the coordinates of Point B into the equation for line AB:

y=x2(3)=(xB)2

Solving, we get:

xB=1
NOTE:

You could also solve this question using the gradient formula. The gradient of line AB is −1 so we could set up this equation using the coordinates from Points A and B:

m=y2y1x2x11=3(1)xB(1)

If you solve this equation, you will get the same answer that we got using the equation of line AB.

The x-coordinate of Point B is 1.


Submit your answer as:

Finding the coordinate of a point using parallel lines

The following diagram shows two parallel lines. Point A is at (3;1) and Point B is at (2;y). Line AB runs parallel to the following line: y=x2, which is the dashed blue line.

Determine the y-coordinate of Point B.

Answer: Point B = (2;).
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the fact that parallel lines have equal gradients.


STEP: Start building the equation for line AB
[−1 point ⇒ 2 / 3 points left]

We need to find the y-coordinate of Point B. We do not know much about line AB. But we do know that it is parallel to the line y=x2. That means the lines have the same gradient! We can read the gradient from the equation.

y=x2gradient=1

Line AB has the same gradient, so we can start building the equation for the line:

y=mx+cyAB=1x+c=x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

We know the coordinates of Point A. We can use them with this equation to calculate the value of c.

y=x+c(1)=1(3)+cc=2

Great! Now we have the equation for line AB:yAB=x+2.


STEP: Substitute the coordinates of Point B into the equation and solve for yB
[−1 point ⇒ 0 / 3 points left]

Finally, we substitute the coordinates of Point B into the equation for line AB:

y=x+2(yB)=(2)+2

Solving, we get:

yB=4
NOTE:

You could also solve this question using the gradient formula. The gradient of line AB is 1 so we could set up this equation using the coordinates from Points A and B:

m=y2y1x2x11=yB(1)2(3)

If you solve this equation, you will get the same answer that we got using the equation of line AB.

The y-coordinate of Point B is 4.


Submit your answer as:

Finding the coordinate of a point using parallel lines

The following diagram shows two parallel lines. Point A is at (3;52) and Point B is at (2;y). Line AB runs parallel to the following line: y=x72, which is the dashed blue line.

Determine the y-coordinate of Point B.

Answer: Point B = (2;).
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the fact that parallel lines have equal gradients.


STEP: Start building the equation for line AB
[−1 point ⇒ 2 / 3 points left]

We need to find the y-coordinate of Point B. We do not know much about line AB. But we do know that it is parallel to the line y=x72. That means the lines have the same gradient! We can read the gradient from the equation.

y=x72gradient=1

Line AB has the same gradient, so we can start building the equation for the line:

y=mx+cyAB=1x+c=x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

We know the coordinates of Point A. We can use them with this equation to calculate the value of c.

y=x+c(52)=1(3)+cc=12

Great! Now we have the equation for line AB:yAB=x12.


STEP: Substitute the coordinates of Point B into the equation and solve for yB
[−1 point ⇒ 0 / 3 points left]

Finally, we substitute the coordinates of Point B into the equation for line AB:

y=x12(yB)=(2)12

Solving, we get:

yB=52
NOTE:

You could also solve this question using the gradient formula. The gradient of line AB is −1 so we could set up this equation using the coordinates from Points A and B:

m=y2y1x2x11=yB522(3)

If you solve this equation, you will get the same answer that we got using the equation of line AB.

The y-coordinate of Point B is 52.


Submit your answer as:

Working with parallel lines

The diagram below shows two parallel lines. The dashed red line has the equation:

ydashed=x3+1

The solid blue line passes through the points A(5;y) and B(8;143), as shown. Note that the y-coordinate of point A is unknown.

Determine yA, the y-coordinate of point A.

Answer: yA=
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the fact that the two lines must have equal gradients because they are parallel. So you can find the gradient of line AB by looking at the equation for the other line.


STEP: Use the fact that parallel lines have equal gradients
[−1 point ⇒ 3 / 4 points left]

The two lines in the graph are parallel. That means they must have equal gradients. The equation for the red dashed line is:

y=x3+1

which means that both of the lines have a gradient of 13. So the equation of line AB must be:

y=x3+c

STEP: Find the value of c for the line using point B
[−1 point ⇒ 2 / 4 points left]

In this question we need a missing coordinate from point A. However, we know both coordinates for point B, so we can use them to find the value of c in the equation for line AB.

y=x3+c(143)=(8)3+cof point B to find cUse the coordinates2=c

Now we have the complete equation for line AB:

y=x32

STEP: Use the x-coordinate of point A to find yA
[−2 points ⇒ 0 / 4 points left]

Now we can substitute the x-coordinate for point A into the equation for line AB and solve for yA. This works because the equation of any line agrees with all of the points on the line: if we substitute in the x-coordinate then when we solve for y we will get the coordinate for the point which belongs on the line: the equation forces this to be true!

yA=xA32yA=(5)32yA=13

The correct answer is yA=13.


Submit your answer as:

Working with parallel lines

The diagram below shows two parallel lines. The dashed red line has the equation:

ydashed=x3+6

The solid blue line passes through the points A(x;23) and B(6;3), as shown. Note that the x-coordinate of point A is unknown.

Determine xA, the x-coordinate of point A.

Answer: xA=
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the fact that the two lines must have equal gradients because they are parallel. So you can find the gradient of line AB by looking at the equation for the other line.


STEP: Use the fact that parallel lines have equal gradients
[−1 point ⇒ 3 / 4 points left]

The two lines in the graph are parallel. That means they must have equal gradients. The equation for the red dashed line is:

y=x3+6

which means that both of the lines have a gradient of 13. So the equation of line AB must be:

y=x3+c

STEP: Find the value of c for the line using point B
[−1 point ⇒ 2 / 4 points left]

In this question we need a missing coordinate from point A. However, we know both coordinates for point B, so we can use them to find the value of c in the equation for line AB.

y=x3+c(3)=(6)3+cof point B to find cUse the coordinates1=c

Now we have the complete equation for line AB:

y=x3+1

STEP: Use the y-coordinate of point A to find xA
[−2 points ⇒ 0 / 4 points left]

Now we can substitute the y-coordinate for point A into the equation for line AB and solve for xA. This works because the equation of any line agrees with all of the points on the line: if we substitute in the y-coordinate then when we solve for x we will get the coordinate for the point which belongs on the line: the equation forces this to be true!

yA=xA3+1(23)=xA3+1xA=5

The correct answer is xA=5.


Submit your answer as:

Working with parallel lines

The diagram below shows two parallel lines. The dashed red line has the equation:

ydashed=x+52

The solid blue line passes through the points A(6;132) and B(2;y), as shown. Note that the y-coordinate of point B is unknown.

Calculate yB, the y-coordinate of point B.

Answer: yB=
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the fact that the two lines must have equal gradients because they are parallel. So you can find the gradient of line AB by looking at the equation for the other line.


STEP: Use the fact that parallel lines have equal gradients
[−1 point ⇒ 3 / 4 points left]

The two lines in the graph are parallel. That means they must have equal gradients. The equation for the red dashed line is:

y=x+52

which means that both of the lines have a gradient of 1. So the equation of line AB must be:

y=x+c

STEP: Find the value of c for the line using point A
[−1 point ⇒ 2 / 4 points left]

In this question we need a missing coordinate from point B. However, we know both coordinates for point A, so we can use them to find the value of c in the equation for line AB.

y=x+c(132)=(6)+cof point A to find cUse the coordinates12=c

Now we have the complete equation for line AB:

y=x12

STEP: Use the x-coordinate of point B to find yB
[−2 points ⇒ 0 / 4 points left]

Now we can substitute the x-coordinate for point B into the equation for line AB and solve for yB. This works because the equation of any line agrees with all of the points on the line: if we substitute in the x-coordinate then when we solve for y we will get the coordinate for the point which belongs on the line: the equation forces this to be true!

yB=xB12yB=(2)12yB=32

The correct answer is yB=32.


Submit your answer as:

Finding the equation of a straight line

The diagram below shows a line. The line has a gradient m=12 and it passes through the point A(7;132).

Find the equation of the line. Give only the right side of the equation, following "y= " below.

Answer: Equation of the line: y =
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You should start with the standard form for a straight line (y=mx+c) and substitute in the gradient. Then use the coordinates of point A to find c.


STEP: Start building the equation using the gradient
[−1 point ⇒ 1 / 2 points left]

The first thing to remember is that a straight line has this general form:

y=mx+c

where m is the gradient and c is the y-intercept. We have the gradient m of the line (it is given in the question) so we can substitute that into the equation of the line:

y=(12)x+cy=x2+c

STEP: Use the coordinates of the point to find c
[−1 point ⇒ 0 / 2 points left]

Now we need the value of c for the line. We can find this by substituing the coordinates of any point into the general form for a straight line. This works because every point on the line solves the equation (the coordinates of the points agree with the equation). Since we have the coordinates of point A we will use them.

y=x2+c(132)=(7)2+c132=72+c3=c

Notice that c is positive: this agrees with the diagram, because the y-intercept is above the x-axis.

The equation of the line is:

y=x2+3

Submit your answer as:

Finding the equation of a straight line

The diagram below shows a line. The line has a gradient m=12 and it passes through the point A(5;2).

Find the equation of the line. Give only the right side of the equation, following "y= " below.

Answer: Equation of the line: y =
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You should start with the standard form for a straight line (y=mx+c) and substitute in the gradient. Then use the coordinates of point A to find c.


STEP: Start building the equation using the gradient
[−1 point ⇒ 1 / 2 points left]

The first thing to remember is that a straight line has this general form:

y=mx+c

where m is the gradient and c is the y-intercept. We have the gradient m of the line (it is given in the question) so we can substitute that into the equation of the line:

y=(12)x+cy=x2+c

STEP: Use the coordinates of the point to find c
[−1 point ⇒ 0 / 2 points left]

Now we need the value of c for the line. We can find this by substituing the coordinates of any point into the general form for a straight line. This works because every point on the line solves the equation (the coordinates of the points agree with the equation). Since we have the coordinates of point A we will use them.

y=x2+c(2)=(5)2+c2=52+c12=c

Notice that c is negative: this agrees with the diagram, because the y-intercept is below the x-axis.

The equation of the line is:

y=x212

Submit your answer as:

Finding the equation of a straight line

The diagram below shows a line. The line has a gradient m=13 and it passes through the point A(7;133).

Calculate the equation of the line. Give only the right side of the equation, following "y= " below.

Answer: Equation of the line: y =
expression
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You should start with the standard form for a straight line (y=mx+c) and substitute in the gradient. Then use the coordinates of point A to find c.


STEP: Start building the equation using the gradient
[−1 point ⇒ 1 / 2 points left]

The first thing to remember is that a straight line has this general form:

y=mx+c

where m is the gradient and c is the y-intercept. We have the gradient m of the line (it is given in the question) so we can substitute that into the equation of the line:

y=(13)x+cy=x3+c

STEP: Use the coordinates of the point to find c
[−1 point ⇒ 0 / 2 points left]

Now we need the value of c for the line. We can find this by substituing the coordinates of any point into the general form for a straight line. This works because every point on the line solves the equation (the coordinates of the points agree with the equation). Since we have the coordinates of point A we will use them.

y=x3+c(133)=(7)3+c133=73+c2=c

Notice that c is negative: this agrees with the diagram, because the y-intercept is below the x-axis.

The equation of the line is:

y=x32

Submit your answer as:

Finding a point on a line

You are given the following diagram:

Point C lies on the line AB.

Calculate the missing coordinate of Point C.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point C = (;2,2).
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the equation of the straight line AB.


STEP: Calculate the gradient (m) of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the unknown variable on the line, we start by calculating the equation of the straight line. We first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(1)(3)(1)(1)m=2

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting Point A into the general form for a straight line:

y=mx+c(3)=(2)×(1)+cc=1

STEP: Solve for x using the known value of Point C
[−1 point ⇒ 0 / 3 points left]

We now substitute the known values of Point C into the known equation of the straight line:

y=mx+cy=(2)x+12,2=(2)(x)+1x=0,6

Submit your answer as:

Finding a point on a line

You are given the following diagram:

Point C lies on the line AB.

Calculate the missing coordinate of Point C.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point C = (2,2;).
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the equation of the straight line AB.


STEP: Calculate the gradient (m) of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the unknown variable on the line, we start by calculating the equation of the straight line. We first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(3,5)(0,5)(3)(5)m=0,5

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting Point A into the general form for a straight line:

y=mx+c(0,5)=(0,5)×(5)+cc=2

STEP: Solve for y using the known value of Point C
[−1 point ⇒ 0 / 3 points left]

We now substitute the known values of Point C into the known equation of the straight line:

y=mx+cy=(0,5)x2y=(0,5)(2,2)2y=3,1

Submit your answer as:

Finding a point on a line

You are given the following diagram:

Point C lies on the line AB.

Calculate the missing coordinate of Point C.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point C = (;1,15).
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by calculating the equation of the straight line AB.


STEP: Calculate the gradient (m) of the line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the unknown variable on the line, we start by calculating the equation of the straight line. We first calculate the gradient (m) of the line AB:

m=yByAxBxAm=(4)(3,5)(2)(3)m=1,5

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting Point A into the general form for a straight line:

y=mx+c(3,5)=(1,5)×(3)+cc=1

STEP: Solve for x using the known value of Point C
[−1 point ⇒ 0 / 3 points left]

We now substitute the known values of Point C into the known equation of the straight line:

y=mx+cy=(1,5)x+11,15=(1,5)(x)+1x=0,1

Submit your answer as:

Finding a coordinate of a point using an equation

You are given the following diagram. It shows point A at (−4;−2) and point B at (x;3).

The equation of line segment AB is: y=x+2

Find the x coordinate of point B, correct to 2 decimal places.

Answer: Point B = (;3)
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the equation together with the coordinates of point B.


STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 2 / 3 points left]

We have two points on a line, but one of the coordinates is missing. We need to find that coordinate. The equation given in the question is the key to finding the answer.

y=x+2

Every point on the line must agree with this equation. That includes the coordinates of point B. We substitute the known value for point B into the equation because that will give us the value of the missing coordinate x which agrees with the equation.

y=x+2(3)=x+2

STEP: Solve the equation for the value of x
[−2 points ⇒ 0 / 3 points left]

Solving, we get:

x=1

Submit your answer as:

Finding a coordinate of a point using an equation

You are given the following diagram. It shows point A at (−1;1) and point B at (x;3,5).

The equation of line segment AB is: y=1,5x0,5

Determine the x coordinate of point B, correct to 2 decimal places.

Answer: Point B = (;3,5)
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the equation together with the coordinates of point B.


STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 2 / 3 points left]

We have two points on a line, but one of the coordinates is missing. We need to find that coordinate. The equation given in the question is the key to finding the answer.

y=−1,5x−0,5

Every point on the line must agree with this equation. That includes the coordinates of point B. We substitute the known value for point B into the equation because that will give us the value of the missing coordinate x which agrees with the equation.

y=1,5x0,5(3,5)=1,5x0,5

STEP: Solve the equation for the value of x
[−2 points ⇒ 0 / 3 points left]

Solving, we get:

x=2

Submit your answer as:

Finding a coordinate of a point using an equation

You are given the following diagram. It shows point A at (−2;0) and point B at (1;y).

The equation of line segment AB is: y=x2

Determine the y coordinate of point B, correct to 2 decimal places.

Answer: Point B = (1;)
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the equation together with the coordinates of point B.


STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 2 / 3 points left]

We have two points on a line, but one of the coordinates is missing. We need to find that coordinate. The equation given in the question is the key to finding the answer.

y=x−2

Every point on the line must agree with this equation. That includes the coordinates of point B. We substitute the known value for point B into the equation because that will give us the value of the missing coordinate y which agrees with the equation.

y=x2y=(1)2

STEP: Solve the equation for the value of y
[−2 points ⇒ 0 / 3 points left]

Solving, we get:

y=3

Submit your answer as:

Finding the equation of a parallel line

What is the equation of a line which is parallel to the line x=4 and passes through the point Z(2;1)?

INSTRUCTION: Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.
Answer:
equation
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. This is a special situation, because the equation given represents a vertical line (equations of the form 'x= a number' are always vertical lines). The gradient of a vertical line is undefined.

For a line which is parallel to this one, you will get another line with an undefined gradient.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Vertical lines always have equations which look like 'x= a number.' In order to find that number, we need to use the x-coordinate from the point given in the equation, (2;1), because we know that the line must pass through that point.

x=a numberx=(2)

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is x=2.

The graph below shows the two equations: the blue line is the equation given in the question, and the red line is the answer. The point Z(2;1) is shown with a black dot and you can see that the red line passes through the point. You can also see that both of the lines are vertical.


Submit your answer as:

Finding the equation of a parallel line

Find the equation of a line passing through the point B(2;4) which is parallel to y=6x1.

INSTRUCTION: Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.
Answer:
equation
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. You can take the gradient of the first line straight from the equation: y=6x1. The gradient for this line is m=6.

For a line which is parallel to this one, the gradient of the line we want is also m=6.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now use the point (2;4), given together with the gradient, to get the equation of the line. Substitute the x and y-values into the equation to get the value of c:

y=mx+c4=(6)(2)+c4=12+c16=c

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is y=6x16.

The graph below shows the two equations: the blue line is the equation given in the question, and the red line is the answer. The point B(2;4) is shown with a black dot and you can see that the red line passes through the point. You can also see that the two lines are parallel.


Submit your answer as:

Finding the equation of a parallel line

What is the equation of a line which is parallel to the line y=x21 and passes through the point B(3;3)?

INSTRUCTION: Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.
Answer:
equation
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. You can take the gradient of the first line straight from the equation: y=x21. The gradient for this line is m=12.

For a line which is parallel to this one, the gradient of the line we want is also m=12.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now use the point (3;3), given together with the gradient, to get the equation of the line. Substitute the x and y-values into the equation to get the value of c:

y=mx+c3=(12)(3)+c3=32+c32=c

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is y=x2+32.

The graph below shows the two equations: the blue line is the equation given in the question, and the red line is the answer. The point B(3;3) is shown with a black dot and you can see that the red line passes through the point. You can also see that the two lines are parallel.


Submit your answer as:

Finding the equation of a straight line

The diagram below shows a straight line. The line passes through the coordinates A(2;52) and B(8;0). Calculate the equation of the line AB.

INSTRUCTION: Type the right side of the equation in the form mx+c.
Answer: y =
expression
STEP: Find the gradient of the line
[−1 point ⇒ 1 / 2 points left]

To find the equation of any straight line, we need the gradient of the line. We have two points on the line, so we can use the gradient formula.

m=yByAxBxA=(0)(52)(8)(2)=14

So the gradient of the line is 14.


STEP: Find the y-intercept of the line
[−1 point ⇒ 0 / 2 points left]

Now we need to find the y-intercept of the line. We do this by substituting the gradient and any point on the line into the general form for a straight line. Here we will use the coordinates of point A. (You can use point B if you want: any point on the line will work.)

y=mx+c(52)=(14)(2)+c52=12+c2=c

Therefore, the equation of the line AB is as follows:

y=x42

Submit your answer as:

Finding the equation of a straight line

The diagram below shows a straight line. The line passes through the coordinates A(2;6) and B(2;6). Find the equation of the line AB.

INSTRUCTION: Type the right side of the equation in the form mx+c.
Answer: y =
expression
STEP: Find the gradient of the line
[−1 point ⇒ 1 / 2 points left]

To find the equation of any straight line, we need the gradient of the line. We have two points on the line, so we can use the gradient formula.

m=yByAxBxA=(6)(6)(2)(2)=3

So the gradient of the line is 3.


STEP: Find the y-intercept of the line
[−1 point ⇒ 0 / 2 points left]

Now we need to find the y-intercept of the line. We do this by substituting the gradient and any point on the line into the general form for a straight line. Here we will use the coordinates of point A. (You can use point B if you want: any point on the line will work.)

y=mx+c(6)=(3)(2)+c6=6+c0=c

Therefore, the equation of the line AB is as follows:

y=3x

Submit your answer as:

Finding the equation of a straight line

The diagram below shows a straight line. The line passes through the coordinates A(4;12) and B(1;34). Determine the equation of the line AB.

INSTRUCTION: Type the right side of the equation in the form mx+c.
Answer: y =
expression
STEP: Find the gradient of the line
[−1 point ⇒ 1 / 2 points left]

To find the equation of any straight line, we need the gradient of the line. We have two points on the line, so we can use the gradient formula.

m=yByAxBxA=(34)(12)(1)(4)=14

So the gradient of the line is 14.


STEP: Find the y-intercept of the line
[−1 point ⇒ 0 / 2 points left]

Now we need to find the y-intercept of the line. We do this by substituting the gradient and any point on the line into the general form for a straight line. Here we will use the coordinates of point A. (You can use point B if you want: any point on the line will work.)

y=mx+c(12)=(14)(4)+c12=1+c12=c

Therefore, the equation of the line AB is as follows:

y=x4+12

Submit your answer as:

Calculating the coordinates of a point on a line

Points A(2;1,5) and B(x;2) lie on the straight line y=0,5x0,5.

Find the x-coordinate of point B, rounded to 2 decimal places.

Answer: Point B = (;2)
numeric
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

We are given the equation of the straight line:

y=0,5x0,5

To determine the value of x, we substitute the y-coordinate for point B into the equation:

y=0,5x0,52=(0,5)(x)0,5

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We rearrange the equation to get the answer:

x=5

Submit your answer as:

Calculating the coordinates of a point on a line

Points A(4;0) and B(x;3,5) lie on the straight line y=0,5x+2.

Determine the x-coordinate of point B, rounded to 2 decimal places.

Answer: Point B = (;3,5)
numeric
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

We are given the equation of the straight line:

y=0,5x+2

To determine the value of x, we substitute the y-coordinate for point B into the equation:

y=0,5x+23,5=(0,5)(x)+2

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We rearrange the equation to get the answer:

x=3

Submit your answer as:

Calculating the coordinates of a point on a line

Points A(2;3) and B(1;y) lie on the straight line y=2x+1.

Calculate the y-coordinate of point B, rounded to 2 decimal places.

Answer: Point B = (1;)
numeric
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

We are given the equation of the straight line:

y=2x+1

To determine the value of y, we substitute the x-coordinate for point B into the equation:

y=2x+1y=(2)(1)+1

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

We rearrange the equation to get the answer:

y=3

Submit your answer as:

Identifying parallel lines

Determine whether or not the following lines are parallel.

2y=53y=1

Note: In tests and exams you must show calculations to support your answer.

Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, you need to know the gradient of each equation. That means that must start by arranging the equations to be in standard form: y=mx+c. If the equations do not contain the variable y, then you need to isolate x to get an equation which says, "x= a number."
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

You can only answer this question by comparing the gradients of the two equations, which means you must arrange the equations to be in standard form. For the first equation this is:

2y=512(2y)=12(5)y=52

This equation is the same as y=0x+52. Therefore, the gradient of this equation is m=0.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now change the second equation to standard form. This will similar to the work above, because there is no x in the equation.

3y=113(3y)=13(1)y=13

As above, we must imagine an x-term in the equation, which will have a coefficient of zero: y=0x13. The gradient of the second equation is also m=0.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Compare the two gradients to decide if the lines are parallel or not. For both of the equations the gradient is zero, so the lines are parallel. The correct choice is: parallel.

The two equations are graphed on the Cartesian plane shown below. You can see that the lines are parallel. Remember that a gradient of zero means that the line must be horizontal.


Submit your answer as:

Identifying parallel lines

Are the following lines parallel?

6x+4y+6=023y=16x+3y15

Note: In tests and exams you must show calculations to support your answer.

Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, you need to know the gradient of each equation. That means that must start by arranging the equations to be in standard form: y=mx+c. If the equations do not contain the variable y, then you need to isolate x to get an equation which says, "x= a number."
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

You can only answer this question by comparing the gradients of the two equations, which means you must arrange the equations to be in standard form. For the first equation this is:

6x+4y+6=04y=6x614(4y)=14(6x6)y=3x232

The gradient of the first equation is m=32.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now change the second equation to standard form.

23y=16x+3y1520y=16x15120(20y)=120(16x15)y=4x534

The gradient of the second equation is m=45.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Compare the two gradients to decide if the lines are parallel or not. For the equations in this question, the gradients are not equal, so the lines cannot be parallel. The correct response is: not parallel.

The two equations are graphed on the Cartesian plane shown below. You can see that the lines are not parallel.


Submit your answer as:

Identifying parallel lines

Are the following lines parallel?

6x+3y+1=0x+2y2=0

Note: In tests and exams you must show calculations to support your answer.

Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, you need to know the gradient of each equation. That means that must start by arranging the equations to be in standard form: y=mx+c. If the equations do not contain the variable y, then you need to isolate x to get an equation which says, "x= a number."
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

You can only answer this question by comparing the gradients of the two equations, which means you must arrange the equations to be in standard form. For the first equation this is:

6x+3y+1=03y=6x113(3y)=13(6x1)y=2x13

The gradient of the first equation is m=2.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now change the second equation to standard form.

x+2y2=02y=x+212(2y)=12(x+2)y=x2+1

The gradient of the second equation is m=12.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Compare the two gradients to decide if the lines are parallel or not. For the equations in this question, the gradients are not equal, so the lines cannot be parallel. The correct response is: not parallel.

The two equations are graphed on the Cartesian plane shown below. You can see that the lines are not parallel.


Submit your answer as:

Calculating a coordinate for a point using the gradient

Look at the following diagram. Point A is at (−2;0,5) and Point B is at (x;3,5). Line AB has a gradient (m) of −1.

Determine the x-coordinate of Point B.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point B= (;-3.5)
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the equation of a straight line, which includes the gradient. Substitute in values that you know.


STEP: Find the equation of the line segment
[−2 points ⇒ 2 / 4 points left]

We have two endpoints of a line segment, and the gradient of the segment. However, the x-coordinate of Point B is missing. We need to find that coordinate value.

We can solve this question using the equation for a straight line:

y=mx+c

Firstly, we can calculate the value of the y-intercept (c) of the line AB. We do this by substituting the gradient value and the coordinates of Point A into the general form for a straight line:

y=mx+c(0,5)=(1)×(2)+cc=1,5

This means that the equation for the line segment in the graph is:

y=x1,5

STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known value for Point B into the equation:

y=x1,5(3,5)=x1,5

STEP: Solve the equation for x
[−1 point ⇒ 0 / 4 points left]

Solving, we get:

x=2

Submit your answer as:

Calculating a coordinate for a point using the gradient

Consider the following diagram. Point A is at (−1;−2,5) and Point B is at (x;1,5). Line AB has a gradient (m) of 2.

Find the x-coordinate of Point B.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point B= (;1.5)
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the equation of a straight line, which includes the gradient. Substitute in values that you know.


STEP: Find the equation of the line segment
[−2 points ⇒ 2 / 4 points left]

We have two endpoints of a line segment, and the gradient of the segment. However, the x-coordinate of Point B is missing. We need to find that coordinate value.

We can solve this question using the equation for a straight line:

y=mx+c

Firstly, we can calculate the value of the y-intercept (c) of the line AB. We do this by substituting the gradient value and the coordinates of Point A into the general form for a straight line:

y=mx+c(2,5)=(2)×(1)+cc=0,5

This means that the equation for the line segment in the graph is:

y=2x0,5

STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known value for Point B into the equation:

y=2x0,5(1,5)=2x0,5

STEP: Solve the equation for x
[−1 point ⇒ 0 / 4 points left]

Solving, we get:

x=1

Submit your answer as:

Calculating a coordinate for a point using the gradient

Examine the following diagram. Point A is at (−4;3) and Point B is at (x;1). Line AB has a gradient (m) of −0,5.

Find the x-coordinate of Point B.

INSTRUCTION: Round your answer to two decimal places.
Answer: Point B= (;-1)
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the equation of a straight line, which includes the gradient. Substitute in values that you know.


STEP: Find the equation of the line segment
[−2 points ⇒ 2 / 4 points left]

We have two endpoints of a line segment, and the gradient of the segment. However, the x-coordinate of Point B is missing. We need to find that coordinate value.

We can solve this question using the equation for a straight line:

y=mx+c

Firstly, we can calculate the value of the y-intercept (c) of the line AB. We do this by substituting the gradient value and the coordinates of Point A into the general form for a straight line:

y=mx+c(3)=(0,5)×(4)+cc=1

This means that the equation for the line segment in the graph is:

y=0,5x+1

STEP: Substitute the coordinates of B into the equation
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known value for Point B into the equation:

y=0,5x+1(1)=0,5x+1

STEP: Solve the equation for x
[−1 point ⇒ 0 / 4 points left]

Solving, we get:

x=4

Submit your answer as:

Finding an equation with parallel lines

You are given the following diagram:

You are also told that line segment AB runs parallel to the following line: y=1,5x3,5. Point A is at (1;2). Determine the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by recalling that the gradient of parallel lines is equal.


STEP: Determine the gradient of line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of the line AB. We can get the gradient from the parallel line. The parallel line has (by definition) the same gradient as the line AB:

y=mx+cy=(1,5)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(2)=(1,5)×(1)+cc=0,5

STEP: Write down the equation of the line AB in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=1,5x+0,5

Submit your answer as:

Finding an equation with parallel lines

You are given the following diagram:

You are also told that line segment AB runs parallel to the following line: y=0,5x+3,5. Point A is at (5;2). Work out the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by recalling that the gradient of parallel lines is equal.


STEP: Determine the gradient of line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of the line AB. We can get the gradient from the parallel line. The parallel line has (by definition) the same gradient as the line AB:

y=mx+cy=(0,5)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(2)=(0,5)×(5)+cc=0,5

STEP: Write down the equation of the line AB in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=0,5x0,5

Submit your answer as:

Finding an equation with parallel lines

You are given the following diagram:

You are also told that line segment AB runs parallel to the following line: y=1,5x+4. Point A is at (1;0,5). Work out the equation of the line AB.

Answer: Equation of the line AB: y = .
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by recalling that the gradient of parallel lines is equal.


STEP: Determine the gradient of line AB
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of the line AB. We can get the gradient from the parallel line. The parallel line has (by definition) the same gradient as the line AB:

y=mx+cy=(1,5)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(0,5)=(1,5)×(1)+cc=1

STEP: Write down the equation of the line AB in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=1,5x+1

Submit your answer as:

Find the equation of a straight line

Calculate the equation of the straight line through the points Q(1;43) and R(52;143). Your answer should be exact (no rounding).

INSTRUCTION: Type the entire equation into the answer box.
Answer: The equation is: .
one-of
type(equation)
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Think about the standard form of a straight line equation: is there anything we can substitute into the formula from the given information?
STEP: Substitute the values into the gradient formula
[−2 points ⇒ 3 / 5 points left]

The standard formula of a straight line is:

y=mx+c

where m is the slope of the graph and c is the y-intercept. We are given only two points on the straight line. So we should use those points to try and find the slope and y-intercept of the equation. Gradient is basically determined by the ratio of vertical change over horizontal change, 'rise-over-run'. We are not given the graph, but we can always use the gradient formula:

m=y2y1x2x1

In this specific case we can represent each coordinate as follows:

x1=1x2=52y1=43y2=143

Now substitute these values into the formula to find the gradient.

gradient=(143)(43)(52)(1)=(6)(32)=(6)×(23)=4

STEP: Now determine the c-value
[−2 points ⇒ 1 / 5 points left]

If we look at the standard form of a straight line equation again, we now have:

y=4x+c
Where 4 is the gradient of the graph. Also notice that this value is positive. This means that the straight line is increasing, moving from the bottom upwards. The next step is to solve for the c-value. In other words we need to find where the graph 'cuts' the y-axis. We can use any point on the straight line to substitute into our formula. Here we will use the given point: Q(1;43).
y=4x+c(43)=4(1)+c(43)+(4)=cc=163

The graph 'cuts' the y-axis at y=163.


STEP: Now substitute all the calculated values into the standard formula
[−1 point ⇒ 0 / 5 points left]

The only thing left to do is to give the full equation of the straight line. We have solved all the missing parts that make up the formula of a straight line.

Therefore the equation is

y=4x+163

Submit your answer as:

Find the equation of a straight line

The points A and B have the coordinates (2;203) and (13;7), respectively. Calculate the equation of the straight line through A and B. Your answer should be exact (no rounding).

INSTRUCTION: Type the entire equation into the answer box.
Answer: The equation is: .
one-of
type(equation)
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Think about the standard form of a straight line equation: is there anything we can substitute into the formula from the given information?
STEP: Substitute the values into the gradient formula
[−2 points ⇒ 3 / 5 points left]

The standard formula of a straight line is:

y=mx+c

where m is the slope of the graph and c is the y-intercept. We are given only two points on the straight line. So we should use those points to try and find the gradient and y-intercept of the equation. Gradient is basically determined by the ratio of vertical change over horizontal change, 'rise-over-run'. We are not given the graph, but we can always use the gradient formula:

m=y2y1x2x1

In this specific case we can represent each coordinate as follows:

x1=2x2=13y1=203y2=7

Now substitute these values into the formula to find the gradient.

gradient=(7)(203)(13)2=(413)(73)=(413)×(37)=(417)

STEP: Now determine the c-value
[−2 points ⇒ 1 / 5 points left]

If we look at the standard form of a straight line equation again, we now have:

y=(417)x+c
Where (417) is the gradient of the graph. Also notice that this value is positive. This means that the straight line is increasing, moving from the bottom upwards. The next step is to solve for the c-value. In other words we need to find where the graph 'cuts' the y-axis. We can use any point on the straight line to substitute into our formula. Here we will use the given point: A(2;203).
y=(417)x+c(203)=(417)(2)+c(203)(827)=cc=10621

The graph 'cuts' the y-axis at y=10621.


STEP: Now substitute all the calculated values into the standard formula
[−1 point ⇒ 0 / 5 points left]

The only thing left to do is to give the full equation of the straight line. We have solved all the missing parts that make up the formula of a straight line.

Therefore the equation is

y=417x10621

Submit your answer as:

Find the equation of a straight line

Determine the equation of the straight line through the points Q(194;112) and R(5;112). Your answer should be exact (no rounding).

INSTRUCTION: Type the entire equation into the answer box.
Answer: The equation is: .
one-of
type(equation)
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]
Think about the standard form of a straight line equation: is there anything we can substitute into the formula from the given information?
STEP: Substitute the values into the gradient formula
[−2 points ⇒ 3 / 5 points left]

The standard formula of a straight line is:

y=mx+c

where m is the slope of the graph and c is the y-intercept. We are given only two points on the straight line. So we should use those points to try and find the gradient and y-intercept of the equation. Gradient is basically determined by the ratio of vertical change over horizontal change, 'rise-over-run'. We are not given the graph, but we can always use the gradient formula:

m=y2y1x2x1

In this specific case we can represent each coordinate as follows:

x1=194x2=5y1=112y2=112

Now substitute these values into the formula to find the gradient.

gradient=(112)(112)(5)(194)=0(394)=0

STEP: Now determine the c-value
[−2 points ⇒ 1 / 5 points left]

If we look at the standard form of a straight line equation again, we now have:

y=0x+c
Where 0 is the gradient of the graph. Also notice that this value is zero. This means that the straight line is a horizontal line, with a 'flat' gradient. The next step is to solve for the c-value. In other words we need to find where the graph 'cuts' the y-axis. We can use any point on the straight line to substitute into our formula. Here we will use the given point: Q(194;112).
y=0x+c(112)=0(194)+c(112)(0)=cc=112


STEP: Now substitute all the calculated values into the standard formula
[−1 point ⇒ 0 / 5 points left]

The only thing left to do is to give the full equation of the straight line. We have solved all the missing parts that make up the formula of a straight line.

Therefore the equation is

y=112

Submit your answer as:

3. Distance

The distance formula with surds

What is the distance between these two points?

A(1;1)B(53;53)

Your answer should be exact. (This means the answer should either be a surd, like 'sqrt(15)', or a whole number. Do not round off your answer.)

Answer: d=
expression
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The coordinates given look a bit complex, but focus on what this question is asking about. Start by writing down the formula for what you need to calculate, and substitute the coordinates into it.


STEP: Substitute the values into the distance formula
[−1 point ⇒ 3 / 4 points left]

This is a distance formula question: we need to calculate the distance between the points (1;1) and (53;53). On a graph, the points and the distance that we want look like this:

To begin, we need to substitute the coordinates into the formula. It can be helpful to list the coordinates from the points according to the values we need in the distance formula.

x1=1;y1=1x2=53;y2=53

Now substitute these values into the formula. Use brackets carefully here to keep everything organised. (There will be lots of brackets, so we will change the brackets in the formula into square brackets to distinguish them from the substituted values.)

d=[x2x1]2+[y2y1]2=[(53)(1)]2+[(53)(1)]2

STEP: Simplify the square brackets and square them
[−2 points ⇒ 1 / 4 points left]

Now we follow BODMAS, which means we need to simplify inside the square brackets as much as possible. However, do not change the surds into decimals: often expressions like this simplify after a few steps. Let's see what happens when we expand the square brackets.

Notice that there is a double negative in the first square bracket which changes to addition.

d=[53+1]2+[531]2=[53+1][53+1]+[531][531]=(75+103+1)+(75103+1)=75+1+75+1

That simplified a lot! The key part is the cancellation of the surd terms, 103 and 103. From this point on there are only integer values to work with inside the root.


STEP: Complete the calculation
[−1 point ⇒ 0 / 4 points left]

Now we can finish the calculation. Remember, we cannot evaluate the square root until all of the calculations inside of it are done, because a root acts like brackets on the expression under it.

d=76+76=152

In this case, the expression cannot be simplified any more: it is a surd. The question tells us the answer must exact, so we have the answer. (This surd can be simplied to 238, which is also accepted.)

The distance between the points is: 152.


Submit your answer as:

Getting to know the distance formula

Consider two points: V(4;7) and W(4;10). Suppose that we want to find the distance between these points. The following shows the distance formula followed by the correct substitution of the coordinates.

d=(x2x1)2+(y2y1)2d=(44)2+((10)7)2
  1. In the substitution step above, what does it mean that (x2x1) will be zero?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If (x2x1)=0, then the x coordinates must be the same. What does that mean about the position of the points on the Cartesian plane?


    STEP: Draw a sketch to see the details
    [−1 point ⇒ 0 / 1 points left]

    The question is about the distance formula. There are two coordinate pairs given, and we must choose an answer for this question: what does it mean that one part of the calculation will be equal to zero? This is easiest to see by graphing the points.

    The points are one above the other! This is because they have equal x coordinates, so they sit at the same x-position: the points are both on the line x=4, as shown in the diagram. The points are separated in the vertical direction, but they are not separated in the horizontal direction. Notice that this means the total distance between the points is exactly equal to the vertical distance between the points, as shown on the diagram.

    The graph above gives the best explanation of what (x2x1)=0 means: the zero comes from the fact that the points have the same x coordinate. That means that there is no horizontal distance between the points.

    The correct answer choice is: There is no horizontal distance between the points.


    Submit your answer as:
  2. Complete the calculation to find the distance between the points given in Question 1.

    INSTRUCTION: Round your answer to two decimal points if appropriate.
    Answer: d=
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You can use the distance formula if you want to, but it is not necessary in this case.


    STEP: Read the answer off the diagram (or calculate it)
    [−2 points ⇒ 0 / 2 points left]

    We can use the formula, but if the distance is vertical or horizontal, it is easy to just count the spaces between the points on the diagram above.

    distance=vertical distance=count from V to W=17

    The distance between points V and W is 17.

    We can of course use the formula to get the same answer. Start with the working shown in Question 1 and continue it.

    d=(x2x1)2+(y2y1)2=(44)2+((10)7)2=(0)2+(17)2

    There is the zero, which represents the fact that there is no horizontal distance between the points. With that zero, the first term disappears because 02 is zero, which is exactly why we could ignore it above.

    d=(17)2=289=17

    Notice that the square root and the square effectively cancelled each other, although the square makes sure that the answer we get is positive (distances must be positive).

    The distance between the points is 17.


    Submit your answer as:

Calculating a line length from two points

The following diagram shows two points on the Cartesian plane. Point A is at (−1;−1) and point B is at (2;−4).

Calculate the length of line segment AB, correct to 2 decimal places.

Answer: Length of segment AB is .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−1 point ⇒ 1 / 2 points left]

First we recall the equation for distance:

dAB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula and determine the distance
[−1 point ⇒ 0 / 2 points left]

Substituting into the equation, we get:

length =(2(1))2+(4(1))2=(2+1)2+(4+1)2=(3)2+(3)2=9+9=18=4,24264...

The length of the line segment, rounded to two decimal places, is 4,24.


Submit your answer as:

Finding a coordinate from a distance

The following picture shows two points on the Cartesian plane, A and B.

The line segment AB¯ has a length of 7,2111. Calculate the missing coordinate of point B. Round your answer to the nearest half-integer.

Answer: The coordinates of point B are: (;2,5).
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−0 points ⇒ 3 / 3 points left]

First we recall the equation for distance:

AB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula
[−1 point ⇒ 2 / 3 points left]

Substituting into the equation, we get:

7,2111=(x(3))2+(2,5(3,5))27,2111=(x+3)2+(2,5+3,5)2

STEP: Solve for x
[−1 point ⇒ 1 / 3 points left]

Now we re-arrange, and solve for the value of x:

(7,2111)2=(x+3)2+(2,5+3,5)252=(x+3)2+(2,5+3,5)252=(x+3)2+36(x+3)2=16x+3=±16x=±43x=(1) or (7)

STEP: Interpret the result and write down the answer
[−1 point ⇒ 0 / 3 points left]

We now have a choice between 2 values for x. From the diagram we can see that the appropriate value for this question is 1.


Submit your answer as:

The distance formula: substituting correctly

Consider two points: C(10;6) and D(5;3). We want to find the distance between these points. According to one of your classmates, Daniel, the following work is the beginning of the solution.

d=(x2x1)2+(y2y1)2=((10)(6))2+(53)2

Are Daniel's substitutions into the formula correct or not?

Answer: The correct choice is: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look at the coordinates in the work: are they substituted in the correct places?


STEP: Consider the options and select the correct choice
[−1 point ⇒ 0 / 1 points left]

The question is about substituting coordinates into the distance formula. It gives us two coordinate pairs and shows the substitution into the formula. We must determine if the substitutions are correct or not.

Let's step back and look at what the question is about. The diagram below shows points C and D on the Cartesian plane. The distance formula calculates the straight-line distance between the points, as shown.

Now back to the substitutions in the question: are the numbers in the correct places? Shame, but Daniel's substitutions are not correct because he mixed up the coordinates. In the distance formula, (x1;y1) represents one of the points, and (x2;y2) represents the other point. While it does not matter if you pick point C or D as (x1;y1), the substitutions must be consistent. Daniel's substitutions are not consistent: specifically, the numbers −6 and 5 are swapped with each other (for example, −6 is a y value, but Daniel plugged it into the formula in an x position). The correct substitution can be either of the following:

d=(5(10))2+(3(6))2-- OR --d=((10)5)2+((6)3)2

Both of these calculations lead to the correct distance, which is 306.

The correct answer choice is: No, there is an error.


Submit your answer as:

Exercises

The distance formula with surds

What is the distance between these two points?

A(215;3)B(3;215)

Your answer should be exact. (This means the answer should either be a surd, like 'sqrt(15)', or a whole number. Do not round off your answer.)

Answer: d=
expression
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The coordinates given look a bit complex, but focus on what this question is asking about. Start by writing down the formula for what you need to calculate, and substitute the coordinates into it.


STEP: Substitute the values into the distance formula
[−1 point ⇒ 3 / 4 points left]

This is a distance formula question: we need to calculate the distance between the points (215;3) and (3;215). On a graph, the points and the distance that we want look like this:

To begin, we need to substitute the coordinates into the formula. It can be helpful to list the coordinates from the points according to the values we need in the distance formula.

x1=215;y1=3x2=3;y2=215

Now substitute these values into the formula. Use brackets carefully here to keep everything organised. (There will be lots of brackets, so we will change the brackets in the formula into square brackets to distinguish them from the substituted values.)

d=[x2x1]2+[y2y1]2=[(3)(215)]2+[(215)(3)]2

STEP: Simplify the square brackets and square them
[−2 points ⇒ 1 / 4 points left]

Now we follow BODMAS, which means we need to simplify inside the square brackets as much as possible. However, do not change the surds into decimals: often expressions like this simplify after a few steps. Let's see what happens when we expand the square brackets.

Notice that there is a double negative in the first square bracket which changes to addition.

d=[3+215]2+[2153]2=[3+215][3+215]+[2153][2153]=(91215+60)+(60+1215+9)=9+60+60+9

That simplified a lot! The key part is the cancellation of the surd terms, 1215 and 1215. From this point on there are only integer values to work with inside the root.


STEP: Complete the calculation
[−1 point ⇒ 0 / 4 points left]

Now we can finish the calculation. Remember, we cannot evaluate the square root until all of the calculations inside of it are done, because a root acts like brackets on the expression under it.

d=69+69=138

In this case, the expression cannot be simplified any more: it is a surd. The question tells us the answer must exact, so we have the answer.

The distance between the points is: 138.


Submit your answer as:

The distance formula with surds

Given the two points here, what is the distance between the points?

V(3;3)W(53;53)

Your answer should be exact. (This means the answer should either be a surd, like 'sqrt(15)', or a whole number. Do not round off your answer.)

Answer: d=
expression
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The coordinates given look a bit complex, but focus on what this question is asking about. Start by writing down the formula for what you need to calculate, and substitute the coordinates into it.


STEP: Substitute the values into the distance formula
[−1 point ⇒ 3 / 4 points left]

This is a distance formula question: we need to calculate the distance between the points (3;3) and (53;53). On a graph, the points and the distance that we want look like this:

To begin, we need to substitute the coordinates into the formula. It can be helpful to list the coordinates from the points according to the values we need in the distance formula.

x1=3;y1=3x2=53;y2=53

Now substitute these values into the formula. Use brackets carefully here to keep everything organised. (There will be lots of brackets, so we will change the brackets in the formula into square brackets to distinguish them from the substituted values.)

d=[x2x1]2+[y2y1]2=[(53)(3)]2+[(53)(3)]2

STEP: Simplify the square brackets and square them
[−2 points ⇒ 1 / 4 points left]

Now we follow BODMAS, which means we need to simplify inside the square brackets as much as possible. However, do not change the surds into decimals: often expressions like this simplify after a few steps. Let's see what happens when we expand the square brackets.

Notice that there is a double negative in the second square bracket which changes to addition.

d=[533]2+[53+3]2=[533][533]+[53+3][53+3]=(75303+9)+(75+303+9)=75+9+75+9

That simplified a lot! The key part is the cancellation of the surd terms, 303 and 303. From this point on there are only integer values to work with inside the root.


STEP: Complete the calculation
[−1 point ⇒ 0 / 4 points left]

Now we can finish the calculation. Remember, we cannot evaluate the square root until all of the calculations inside of it are done, because a root acts like brackets on the expression under it.

d=84+84=168

In this case, the expression cannot be simplified any more: it is a surd. The question tells us the answer must exact, so we have the answer. (This surd can be simplied to 242, which is also accepted.)

The distance between the points is: 168.


Submit your answer as:

The distance formula with surds

Calculate the distance between these points:

A(1;52)B(52;1)

Your answer should be exact. (This means the answer should either be a surd, like 'sqrt(15)', or a whole number. Do not round off your answer.)

Answer: d=
expression
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The coordinates given look a bit complex, but focus on what this question is asking about. Start by writing down the formula for what you need to calculate, and substitute the coordinates into it.


STEP: Substitute the values into the distance formula
[−1 point ⇒ 3 / 4 points left]

This is a distance formula question: we need to calculate the distance between the points (1;52) and (52;1). On a graph, the points and the distance that we want look like this:

To begin, we need to substitute the coordinates into the formula. It can be helpful to list the coordinates from the points according to the values we need in the distance formula.

x1=1;y1=52x2=52;y2=1

Now substitute these values into the formula. Use brackets carefully here to keep everything organised. (There will be lots of brackets, so we will change the brackets in the formula into square brackets to distinguish them from the substituted values.)

d=[x2x1]2+[y2y1]2=[(52)(1)]2+[(1)(52)]2

STEP: Simplify the square brackets and square them
[−2 points ⇒ 1 / 4 points left]

Now we follow BODMAS, which means we need to simplify inside the square brackets as much as possible. However, do not change the surds into decimals: often expressions like this simplify after a few steps. Let's see what happens when we expand the square brackets.

Notice that there is a double negative inside both of the square brackets: these both become addition.

d=[52+1]2+[1+52]2=[52+1][52+1]+[1+52][1+52]=(50102+1)+(1+102+50)=50+1+1+50

That simplified a lot! The key part is the cancellation of the surd terms, 102 and 102. From this point on there are only integer values to work with inside the root.


STEP: Complete the calculation
[−1 point ⇒ 0 / 4 points left]

Now we can finish the calculation. Remember, we cannot evaluate the square root until all of the calculations inside of it are done, because a root acts like brackets on the expression under it.

d=51+51=102

In this case, the expression cannot be simplified any more: it is a surd. The question tells us the answer must exact, so we have the answer.

The distance between the points is: 102.


Submit your answer as:

Getting to know the distance formula

Consider two points: P(1;7) and Q(1;4). Suppose that we want to find the distance between these points. The following shows the distance formula followed by the correct substitution of the coordinates.

d=(x2x1)2+(y2y1)2d=(11)2+((7)4)2
  1. In the substitution step above, what does it mean that (x2x1) will be zero?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If (x2x1)=0, then the x coordinates must be the same. What does that mean about the position of the points on the Cartesian plane?


    STEP: Draw a sketch to see the details
    [−1 point ⇒ 0 / 1 points left]

    The question is about the distance formula. There are two coordinate pairs given, and we must choose an answer for this question: what does it mean that one part of the calculation will be equal to zero? This is easiest to see by graphing the points.

    The points are one above the other! This is because they have equal x coordinates, so they sit at the same x-position: the points are both on the line x=1, as shown in the diagram. The points are separated in the vertical direction, but they are not separated in the horizontal direction. Notice that this means the total distance between the points is exactly equal to the vertical distance between the points, as shown on the diagram.

    The graph above gives the best explanation of what (x2x1)=0 means: the zero comes from the fact that the points have the same x coordinate. That means that there is no horizontal distance between the points.

    The correct answer choice is: There is no horizontal distance between the points.


    Submit your answer as:
  2. Complete the calculation to find the distance between the points given in Question 1.

    INSTRUCTION: Round your answer to two decimal points if appropriate.
    Answer: d=
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You can use the distance formula if you want to, but it is not necessary in this case.


    STEP: Read the answer off the diagram (or calculate it)
    [−2 points ⇒ 0 / 2 points left]

    We can use the formula, but if the distance is vertical or horizontal, it is easy to just count the spaces between the points on the diagram above.

    distance=vertical distance=count from P to Q=11

    The distance between points P and Q is 11.

    We can of course use the formula to get the same answer. Start with the working shown in Question 1 and continue it.

    d=(x2x1)2+(y2y1)2=(11)2+((7)4)2=(0)2+(11)2

    There is the zero, which represents the fact that there is no horizontal distance between the points. With that zero, the first term disappears because 02 is zero, which is exactly why we could ignore it above.

    d=(11)2=121=11

    Notice that the square root and the square effectively cancelled each other, although the square makes sure that the answer we get is positive (distances must be positive).

    The distance between the points is 11.


    Submit your answer as:

Getting to know the distance formula

Consider two points: V(5;10) and W(2;4). Suppose that we want to find the distance between these points. The following shows the distance formula followed by the correct substitution of the coordinates.

d=(x2x1)2+(y2y1)2d=((2)5)2+((4)10)2
  1. What is the distance formula based on?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Think of squaring both sides of the distance formula to get this:

    d2=(x2x1)2+(y2y1)2

    What formula do you know which looks similar?


    STEP: Draw a sketch to see the details
    [−1 point ⇒ 0 / 1 points left]

    The question is about the distance formula. There are two coordinate pairs given, but we do not need them for this question. Instead, we need to identify the foundation of the distance formula (what it is based on). This is easiest to see by graphing the points.

    As you can see, we have built a right-angled triangle onto the points. We The distance between the points is the hypotenuse of the triangle. The quantity (x2x1) corresponds to the horizontal leg of the triangle and (y2y1) corresponds to the vertical leg of the triangle. Since this is a right-angled triangle, the theorem of Pythagoras applies to the lengths of the sides. In fact, the distance formula is built on the theorem of Pythagoras!

    The theorem of Pythagoras says that:

    c2=a2+b2

    where c is the hypotenuse of the triangle. The hypotenuse of the triangle above is the distance between the points. The other sides of the triangle correspond to sides a and b in Pythagoras's equation. Substituting the three sides of the triangle in the diagram above into the theorem, we get:

    c2=a2+b2distance2=(x2x1)2+(y2y1)2

    If we square root both sides, we get the distance formula:

    distance=(x2x1)2+(y2y1)2

    The correct answer choice is: the theorem of Pythagoras.


    Submit your answer as:
  2. Complete the calculation to find the distance between the points given in Question 1.

    INSTRUCTION: Round your answer to two decimal points if appropriate.
    Answer: d=
    numeric
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Continue the working started in the answer above.


    STEP: Work out the answer following BODMAS
    [−3 points ⇒ 0 / 3 points left]

    To determine the distance between the points, we need to complete the calculation which was started in Question 1. We already had:

    d=(x2x1)2+(y2y1)2=((2)5)2+((4)10)2

    Remember to follow BODMAS, and that a root acts like brackets (this means you must do all of the calculations under the root before you calculate the square root itself).

    d=(25)2+(410)2=(7)2+(14)2

    We still cannot evaluate the root: we must finish all calculations inside the root first, and then take the square root of that result.

    d=49+196=245=15,65247...15,65

    245 is not a perfect square, so we need to round the decimal to two places (according to the instructions).

    The distance between the points is 15,65.


    Submit your answer as:

Getting to know the distance formula

Consider two points: V(8;10) and W(8;5). Suppose that we want to find the distance between these points. The following shows the distance formula followed by the correct substitution of the coordinates.

d=(x2x1)2+(y2y1)2d=(88)2+((10)5)2
  1. In the substitution step above, what does it mean that (x2x1) will be zero?

    Answer:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    If (x2x1)=0, then the x coordinates must be the same. What does that mean about the position of the points on the Cartesian plane?


    STEP: Draw a sketch to see the details
    [−1 point ⇒ 0 / 1 points left]

    The question is about the distance formula. There are two coordinate pairs given, and we must choose an answer for this question: what does it mean that one part of the calculation will be equal to zero? This is easiest to see by graphing the points.

    The points are one above the other! This is because they have equal x coordinates, so they sit at the same x-position: the points are both on the line x=8, as shown in the diagram. The points are separated in the vertical direction, but they are not separated in the horizontal direction. Notice that this means the total distance between the points is exactly equal to the vertical distance between the points, as shown on the diagram.

    The graph above gives the best explanation of what (x2x1)=0 means: the zero comes from the fact that the points have the same x coordinate. That means that there is no horizontal distance between the points.

    The correct answer choice is: There is no horizontal distance between the points.


    Submit your answer as:
  2. Complete the calculation to find the distance between the points given in Question 1.

    INSTRUCTION: Round your answer to two decimal points if appropriate.
    Answer: d=
    numeric
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    You can use the distance formula if you want to, but it is not necessary in this case.


    STEP: Read the answer off the diagram (or calculate it)
    [−2 points ⇒ 0 / 2 points left]

    We can use the formula, but if the distance is vertical or horizontal, it is easy to just count the spaces between the points on the diagram above.

    distance=vertical distance=count from V to W=15

    The distance between points V and W is 15.

    We can of course use the formula to get the same answer. Start with the working shown in Question 1 and continue it.

    d=(x2x1)2+(y2y1)2=(88)2+((10)5)2=(0)2+(15)2

    There is the zero, which represents the fact that there is no horizontal distance between the points. With that zero, the first term disappears because 02 is zero, which is exactly why we could ignore it above.

    d=(15)2=225=15

    Notice that the square root and the square effectively cancelled each other, although the square makes sure that the answer we get is positive (distances must be positive).

    The distance between the points is 15.


    Submit your answer as:

Calculating a line length from two points

The following diagram shows two points on the Cartesian plane. Point A is at (−1;0) and point B is at (1;−3).

Calculate the length of line segment AB, correct to 2 decimal places.

Answer: Length of segment AB is .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−1 point ⇒ 1 / 2 points left]

First we recall the equation for distance:

dAB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula and determine the distance
[−1 point ⇒ 0 / 2 points left]

Substituting into the equation, we get:

length =(1(1))2+(3(0))2=(1+1)2+(3+0)2=(2)2+(3)2=4+9=13=3,60555...

The length of the line segment, rounded to two decimal places, is 3,61.


Submit your answer as:

Calculating a line length from two points

The following diagram shows two points on the Cartesian plane. Point A is at (−2;−3) and point B is at (1;3).

Calculate the length of line segment AB, correct to 2 decimal places.

Answer: Length of segment AB is .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−1 point ⇒ 1 / 2 points left]

First we recall the equation for distance:

dAB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula and determine the distance
[−1 point ⇒ 0 / 2 points left]

Substituting into the equation, we get:

length =(1(2))2+(3(3))2=(1+2)2+(3+3)2=(3)2+(6)2=9+36=45=6,70820...

The length of the line segment, rounded to two decimal places, is 6,71.


Submit your answer as:

Calculating a line length from two points

The following diagram shows two points on the Cartesian plane. Point A is at (−1;0,5) and point B is at (1;−3,5).

Calculate the length of line segment AB, correct to 2 decimal places.

Answer: Length of segment AB is .
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−1 point ⇒ 1 / 2 points left]

First we recall the equation for distance:

dAB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula and determine the distance
[−1 point ⇒ 0 / 2 points left]

Substituting into the equation, we get:

length =(1(1))2+(3,5(0,5))2=(1+1)2+(3,50,5)2=(2)2+(4)2=4+16=20=4,47213...

The length of the line segment, rounded to two decimal places, is 4,47.


Submit your answer as:

Finding a coordinate from a distance

The following picture shows two points on the Cartesian plane, A and B.

The line segment AB¯ has a length of 3,6056. Calculate the missing coordinate of point B. Round your answer to the nearest half-integer.

Answer: The coordinates of point B are: (1;).
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−0 points ⇒ 3 / 3 points left]

First we recall the equation for distance:

AB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula
[−1 point ⇒ 2 / 3 points left]

Substituting into the equation, we get:

3,6056=(1(1))2+(y(3))23,6056=(1+1)2+(y+3)2

STEP: Solve for y
[−1 point ⇒ 1 / 3 points left]

Now we re-arrange, and solve for the value of y:

(3,6056)2=(1+1)2+(y+3)213=(1+1)2+(y+3)213=(y+3)2+4(y+3)2=9y+3=±9y=±33y=(0) or (6)

STEP: Interpret the result and write down the answer
[−1 point ⇒ 0 / 3 points left]

We now have a choice between 2 values for y. From the diagram we can see that the appropriate value for this question is 0.


Submit your answer as:

Finding a coordinate from a distance

The following picture shows two points on the Cartesian plane, A and B.

The line segment AB¯ has a length of 6,7082. Calculate the missing coordinate of point B. Round your answer to the nearest half-integer.

Answer: The coordinates of point B are: (;4).
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−0 points ⇒ 3 / 3 points left]

First we recall the equation for distance:

AB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula
[−1 point ⇒ 2 / 3 points left]

Substituting into the equation, we get:

6,7082=(x(2))2+(4(2))26,7082=(x+2)2+(42)2

STEP: Solve for x
[−1 point ⇒ 1 / 3 points left]

Now we re-arrange, and solve for the value of x:

(6,7082)2=(x+2)2+(42)245=(x+2)2+(42)245=(x+2)2+36(x+2)2=9x+2=±9x=±32x=(1) or (5)

STEP: Interpret the result and write down the answer
[−1 point ⇒ 0 / 3 points left]

We now have a choice between 2 values for x. From the diagram we can see that the appropriate value for this question is 1.


Submit your answer as:

Finding a coordinate from a distance

The following picture shows two points on the Cartesian plane, A and B.

The line segment AB¯ has a length of 5,6569. Calculate the missing coordinate of point B. Round your answer to the nearest half-integer.

Answer: The coordinates of point B are: (1;).
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the distance formula: AB=(xBxA)2+(yByA)2, where (xA;yA) are the coordinates of point A and (xB;yB) are the coordinates of the point B.


STEP: Recall the equation for distance
[−0 points ⇒ 3 / 3 points left]

First we recall the equation for distance:

AB=(xBxA)2+(yByA)2

STEP: Substitute the coordinates of points A and B into the distance formula
[−1 point ⇒ 2 / 3 points left]

Substituting into the equation, we get:

5,6569=(1(3))2+(y(1))25,6569=(1+3)2+(y1)2

STEP: Solve for y
[−1 point ⇒ 1 / 3 points left]

Now we re-arrange, and solve for the value of y:

(5,6569)2=(1+3)2+(y1)232=(1+3)2+(y1)232=(y1)2+16(y1)2=16y1=±16y=±4+1y=(5) or (3)

STEP: Interpret the result and write down the answer
[−1 point ⇒ 0 / 3 points left]

We now have a choice between 2 values for y. From the diagram we can see that the appropriate value for this question is 3.


Submit your answer as:

The distance formula: substituting correctly

Consider two points: V(3;2) and W(4;9). We want to find the distance between these points. According to one of your classmates, Riaan, the following work is the beginning of the solution.

d=(x2x1)2+(y2y1)2=((3)2)2+(4(9))2

Are Riaan's substitutions into the formula correct or not?

Answer: The correct choice is: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look at the coordinates in the work: are they substituted in the correct places?


STEP: Consider the options and select the correct choice
[−1 point ⇒ 0 / 1 points left]

The question is about substituting coordinates into the distance formula. It gives us two coordinate pairs and shows the substitution into the formula. We must determine if the substitutions are correct or not.

Let's step back and look at what the question is about. The diagram below shows points V and W on the Cartesian plane. The distance formula calculates the straight-line distance between the points, as shown.

Now back to the substitutions in the question: are the numbers in the correct places? Shame, but Riaan's substitutions are not correct because he mixed up the coordinates. In the distance formula, (x1;y1) represents one of the points, and (x2;y2) represents the other point. While it does not matter if you pick point V or W as (x1;y1), the substitutions must be consistent. Riaan's substitutions are not consistent: specifically, the numbers 2 and 4 are swapped with each other (for example, 2 is a y value, but Riaan plugged it into the formula in an x position). The correct substitution can be either of the following:

d=(4(3))2+((9)2)2-- OR --d=((3)4)2+(2(9))2

Both of these calculations lead to the correct distance, which is 170.

The correct answer choice is: No, there is an error.


Submit your answer as:

The distance formula: substituting correctly

Consider two points: C(10;2) and D(8;6). We want to find the distance between these points. According to one of your classmates, Mark, the following work is the beginning of the solution.

d=(x2x1)2+(y2y1)2=(8(10))2+((2)6)2

Are Mark's substitutions into the formula correct or not?

Answer: The correct choice is: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look at the coordinates in the work: are they substituted in the correct places?


STEP: Consider the options and select the correct choice
[−1 point ⇒ 0 / 1 points left]

The question is about substituting coordinates into the distance formula. It gives us two coordinate pairs and shows the substitution into the formula. We must determine if the substitutions are correct or not.

Let's step back and look at what the question is about. The diagram below shows points C and D on the Cartesian plane. The distance formula calculates the straight-line distance between the points, as shown.

Now back to the substitutions in the question: are the numbers in the correct places? Shame, but Mark's substitutions are not correct because he mixed up the coordinates. In the distance formula, (x1;y1) represents one of the points, and (x2;y2) represents the other point. While it does not matter if you pick point C or D as (x1;y1), the substitutions must be consistent. Mark's substitutions are not consistent: the x-coordinates and the y-coordinates are substituted backwards compared to each other. Either the numbers 6 and 2 should be swapped or the numbers 10 and 8 should be swapped. The correct substitution can be either of the following:

d=(8(10))2+(6(2))2-- OR --d=((10)8)2+((2)6)2

Both of these calculations lead to the correct distance, which is 388.

You might notice that Mark's substitutions are a combination of the two correct versions shown above. Because of that, Mark would actually be lucky and get the correct answer. However, this is only because the squares in the formula cancel out his error. Even while he may be lucky to get the correct answer, the substitutions are not fully correct.

The correct answer choice is: No, there is an error.


Submit your answer as:

The distance formula: substituting correctly

Consider two points: P(5;1) and Q(5;6). We want to find the distance between these points. According to one of your classmates, William, the following work is the beginning of the solution.

d=(x2x1)2+(y2y1)2=((5)5)2+(1(6))2

Are William's substitutions into the formula correct or not?

Answer: The correct choice is: .
HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Look at the coordinates in the work: are they substituted in the correct places?


STEP: Consider the options and select the correct choice
[−1 point ⇒ 0 / 1 points left]

The question is about substituting coordinates into the distance formula. It gives us two coordinate pairs and shows the substitution into the formula. We must determine if the substitutions are correct or not.

Let's step back and look at what the question is about. The diagram below shows points P and Q on the Cartesian plane. The distance formula calculates the straight-line distance between the points, as shown.

Now back to the substitutions in the question: are the numbers in the correct places? In this case, yes they are. In the distance formula, (x1;y1) represents one of the points, and (x2;y2) represents the other point. William is using point Q as (x1;y1) and P as (x2;y2). He could do it the other way around, and that would be correct too. In other words, both of the following are correct:

d=(5(5))2+((6)1)2-- OR --d=((5)5)2+(1(6))2

Both of these calculations lead to the correct distance, which is 149.

The correct answer choice is: Yes, the substitutions are correct.


Submit your answer as:

4. Midpoint

Finding the mid-point of two points

The following diagram shows points A(1;1,5) and B(2;1,5). Point M is the mid-point of A and B.

Calculate the coordinates of the mid-point, M.

INSTRUCTION: Remember to type your answer with brackets and with ';' between the coordinates, for example: ( 3/2 ; -4).
Answer: The coordinates of M are: .
coordinate
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Write down the mid-point formula. (If you do not remember it, look it up in your notes or in a book.)
STEP: Write down the mid-point formula
[−1 point ⇒ 2 / 3 points left]

Start with the mid-point formula:

M=(x1+x22;y1+y22)

In this formula, M is the mid-point of two other points. The coordinates in the formula, (x1;y1) and (x2;y2), are the coordinates of those other points.


STEP: Identify the values of x1,x2,y1 and y2 and substitute into the formula
[−1 point ⇒ 1 / 3 points left]

In this question, (x1;y1) and (x2;y2) must be the coordinates of points A and B. Therefore:

x1=1;y1=1,5;x2=2;andy2=1,5

Now substitute these values into the formula:

M=((1)+(2)2;(1,5)+(1,5)2)

STEP: Evaluate the answers
[−1 point ⇒ 0 / 3 points left]

Working out the calculations, we get the final answers:

M=((1)+(2)2;(1,5)+(1,5)2)=(12;02)=(12;0)

We can compare this answer to the diagram above to check if the answer is reasonable: does the position of point M agree with the coordinates (12;0)? Indeed, that looks quite reasonable!

The coordinates of the mid-point are (12;0).


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Finding the mid-point from coordinates

There are two points, A(5;174) and B(5;3). What is the mid-point of A and B? Your answer should be exact (no rounding).

Your answer should be in brackets, with the ';' symbol between the coordinates. For example: ( 1/2 ; -5).

Answer: The mid-point is: .
coordinate
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Use the mid-point formula. Start by figuring out the values of x1,x2,y1 and y2.
STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 3 points left]

We have two points and we need to find their mid-point. For this we must use the mid-point formula:

M=(x1+x22;y1+y22)

In this equation, M represents the mid-point, which has the coordinates (xM;yM), while the variables x1, etc. refer to the coordinates of the end points. In this case, the end points are the points A and B. We must decide which point to use as point 1 and which to use as point 2. It does not matter which way we do it, so we will just use the order in which the coordinates were given in the question:

point 1:(5;174)point 2:(5;3)

This means:

x1=5x2=5y1=174y2=3

Plugging these values into the mid-point formula, we get:

(xM;yM)=(5+(5)2;(174)+32)

STEP: Substitute and evaluate each expression
[−2 points ⇒ 0 / 3 points left]

Now evaluate the expressions to get the answers. Remember that there are two different expressions in the mid-point formula, one for the x coordinates and the other for the y coordinates. It is helpful to separate the calculations, like this:

x coordinatesy coordinatesxM=5+(5)2yM=(174)+32=02=(294)2=0=(294)×12=298

The resulting coordinates for the mid-point are (0;298). Notice that the x coordinate of the mid-point is zero, which happens because the x coordinates of A and B are opposite values.

We can do a simple check to see if this is reasonable by making a quick sketch of the points and the answer. You can see that the coordinates we got do in fact sit directly between points A and B.

The mid-point of A and B is at: (0;298).


Submit your answer as:

Working with algebraic coordinates and the mid-point

Consider two Points R and Q. Determine the coordinates of Q if R is at (p2;2q21) and the mid-point of R and Q is (0;2).

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of Q is .
  2. The y-coordinate of Q is .
expression
expression
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The question is about a mid-point: write down the mid-point formula and think about what values you can substitute into it.


STEP: Organise the information and make a plan
[−1 point ⇒ 3 / 4 points left]

In this question we have information about three points: we know the coordinates of one end point and the coordinates of the mid-points, but we do not know the coordinates of Point Q.

Point Q:(xQ;yQ)Point R:(p2;2q21)the mid-point:(0;2)

We do not know the coordinates of Point Q, so we use variables to represent them.

The coordinates are given in terms of p and q, but do not let the variables distract you! The question is about a mid-point, and that means a point exactly halfway between two other points. In other words, the situation is something like the picture below, and we can solve it through the mid-point formula.

M=(x1+x22;y1+y22)

STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

Substitute in the values we have from above into the mid-point formula. The coordinates we have for the mid-point must go in on the left side.

(xM;yM)=(x1+x22;y1+y22)(0;2)=(xQ+(p2)2;yQ+(2q21)2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The result above represents two separate equations: one for x and another for y. We can write them separately and solve each one. Begin by multiplying the equations by 2 to cancel the denominators.

x-coordinates0=xQ+(p2)22(0)=xQp20=xQp20+p2=xQp2=xQy-coordinates2=yQ+(2q21)22(2)=yQ2q214=yQ2q214+2q2+1=yQ2q2+5=yQ

We cannot simplify the expressions any more, so we are done. The resulting answers are the coordinates for the end Point Q: (p2;2q2+5).

The final answers are:

  1. The x-coordinate of Q is p2.
  2. The y-coordinate of Q is 2q2+5.

Submit your answer as: and

Using a mid-point to find an end point

The mid-point of Point A(2;6) and another Point B is at (4;92). This is represented on the diagram below.

What are the coordinates of Point B?

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of B is .
  2. The y-coordinate of B is .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the question includes a mid-point, think about how you can use the mid-point formula.


STEP: Set up the equations based on the mid-point formula
[−2 points ⇒ 2 / 4 points left]

This question is about a mid-point, which as always is a point exactly between two other points. In this case, however, we only know the coordinates of one of those end points:

Point A:(2;6)Point B:(xB;yB)the mid-point:(4;92)

We do not know the coordinates of Point B, so we use variables to represent them.

Since this question is about a mid-point, and we know the coordinates of some of the points involved, we should approach the question through the mid-point formula:

M=(x1+x22;y1+y22)(xM;yM)=(x1+x22;y1+y22)

In this case, we know that the mid-point coordinates are (4;92), but we do not know the coordinates of Point B. So the equation becomes:

(4;92)=(2+xB2;6+yB2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The equation above actually represents two separate equations: one for x and another for y. Let's write those two equations separately, and solve! Start by cancelling the denominators in the equations. We do this by multiplying both sides of the equation by the LCD of the fractions in the equation. For both equations, this means multiplying by 2.

x-coordinates4=2+xB22(4)=(2+xB2)28=2+xB82=xB6=xBy-coordinates92=6+yB22(92)=(6+yB2)29=6+yB96=yB3=yB

We get the answers xB = 6 and yB = 3. Compare these numbers to the graph in the question: we can see that the answer is consistent with the position of Point B on the graph.

Point B must be at (6;3), which means the final answers are:

  1. The x-coordinate of B is 6.
  2. The y-coordinate of B is 3.

Submit your answer as: and

Mid-point calculations with algebra

The points C and D have the following coordinates:

C(2h2+5;j2+3)D(2h21;3j2+4),h,jN

Calculate the coordinates of the mid-point of C and D. Type your answer with brackets and the ';' symbol between the coordinates. For example, your answer might look like this: ( a + 3 ; -2b ).

Answer:
The mid-point coordinates are: .
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Use the mid-point formula. If the variables seem to make the question complex, try to ignore them and simply think about working with the formula.
STEP: Organise the information in the question
[−1 point ⇒ 3 / 4 points left]

This question is about finding a mid-point. It is complex because the coordinates contain variables, but in the end it is still a mid-point calculation. Start by writing down the information from the question. (Notice that the information "h,jN" does not matter: for the mid-point it is the coordinate values which we need.)

point C:(2h2+5;j2+3)point D:(2h21;3j2+4)the mid-point:(xM;yM)coordinates so we must use variablesWe do not know the mid-point

Remember that this is about a mid-point, which means the situation looks somehow like the picture below. We have no idea where the points actually are, but the situation is something like the picture.


STEP: Substitute into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

We need to substitute the coordinates above into the mid-point formula because we want to find the mid-point of points C and D. The mid-point formula is:

M=(x1+x22;y1+y22)

where x1,y1, etc. represent the coordinates of the end points (C and D). So:

x1=2h2+5,x2=2h21,y1=j2+3,y2=3j2+4

Substitute these expressions into the formula. Make sure you use brackets around the substitutions to help stay organised.

(xM;yM)=((2h2+5)+(2h21)2;(j2+3)+(3j2+4)2)

STEP: Simplify each expression
[−2 points ⇒ 0 / 4 points left]

Now we need to simplify as much as possible. Remember that there are two different expressions to evaluate in the mid-point formula, one for x and one for y. We will separate these calculations to simplify things.

x coordinatesy coordinatesxM=(2h2+5)+(2h21)2yM=(j2+3)+(3j2+4)2=2h2+52h212=j2+3+3j2+42=42=4j2+72=2=4j22+72=2j2+72

There we have the answer! The mid-point of the two points is: (2;2j2+72).


Submit your answer as:

Exercises

Finding the mid-point of two points

The following diagram shows points A(2;3,5) and B(1;2,5). Point M is the mid-point of A and B.

Calculate the coordinates of the mid-point, M.

INSTRUCTION: Remember to type your answer with brackets and with ';' between the coordinates, for example: ( 3/2 ; -4).
Answer: The coordinates of M are: .
coordinate
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Write down the mid-point formula. (If you do not remember it, look it up in your notes or in a book.)
STEP: Write down the mid-point formula
[−1 point ⇒ 2 / 3 points left]

Start with the mid-point formula:

M=(x1+x22;y1+y22)

In this formula, M is the mid-point of two other points. The coordinates in the formula, (x1;y1) and (x2;y2), are the coordinates of those other points.


STEP: Identify the values of x1,x2,y1 and y2 and substitute into the formula
[−1 point ⇒ 1 / 3 points left]

In this question, (x1;y1) and (x2;y2) must be the coordinates of points A and B. Therefore:

x1=2;y1=3,5;x2=1;andy2=2,5

Now substitute these values into the formula:

M=((2)+(1)2;(3,5)+(2,5)2)

STEP: Evaluate the answers
[−1 point ⇒ 0 / 3 points left]

Working out the calculations, we get the final answers:

M=((2)+(1)2;(3,5)+(2,5)2)=(12;12)

We can compare this answer to the diagram above to check if the answer is reasonable: does the position of point M agree with the coordinates (12;12)? Indeed, that looks quite reasonable!

The coordinates of the mid-point are (12;12).


Submit your answer as:

Finding the mid-point of two points

The following diagram shows points A(2;3) and B(1;0). Point M is the mid-point of A and B.

Calculate the coordinates of the mid-point, M.

INSTRUCTION: Remember to type your answer with brackets and with ';' between the coordinates, for example: ( 3/2 ; -4).
Answer: The coordinates of M are: .
coordinate
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Write down the mid-point formula. (If you do not remember it, look it up in your notes or in a book.)
STEP: Write down the mid-point formula
[−1 point ⇒ 2 / 3 points left]

Start with the mid-point formula:

M=(x1+x22;y1+y22)

In this formula, M is the mid-point of two other points. The coordinates in the formula, (x1;y1) and (x2;y2), are the coordinates of those other points.


STEP: Identify the values of x1,x2,y1 and y2 and substitute into the formula
[−1 point ⇒ 1 / 3 points left]

In this question, (x1;y1) and (x2;y2) must be the coordinates of points A and B. Therefore:

x1=2;y1=3;x2=1;andy2=0

Now substitute these values into the formula:

M=((2)+(1)2;(3)+(0)2)

STEP: Evaluate the answers
[−1 point ⇒ 0 / 3 points left]

Working out the calculations, we get the final answers:

M=((2)+(1)2;(3)+(0)2)=(12;32)

We can compare this answer to the diagram above to check if the answer is reasonable: does the position of point M agree with the coordinates (12;32)? Indeed, that looks quite reasonable!

The coordinates of the mid-point are (12;32).


Submit your answer as:

Finding the mid-point of two points

The following diagram shows points A(3;4) and B(1;4). Point M is the mid-point of A and B.

Calculate the coordinates of the mid-point, M.

INSTRUCTION: Remember to type your answer with brackets and with ';' between the coordinates, for example: ( 3/2 ; -4).
Answer: The coordinates of M are: .
coordinate
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Write down the mid-point formula. (If you do not remember it, look it up in your notes or in a book.)
STEP: Write down the mid-point formula
[−1 point ⇒ 2 / 3 points left]

Start with the mid-point formula:

M=(x1+x22;y1+y22)

In this formula, M is the mid-point of two other points. The coordinates in the formula, (x1;y1) and (x2;y2), are the coordinates of those other points.


STEP: Identify the values of x1,x2,y1 and y2 and substitute into the formula
[−1 point ⇒ 1 / 3 points left]

In this question, (x1;y1) and (x2;y2) must be the coordinates of points A and B. Therefore:

x1=3;y1=4;x2=1;andy2=4

Now substitute these values into the formula:

M=((3)+(1)2;(4)+(4)2)

STEP: Evaluate the answers
[−1 point ⇒ 0 / 3 points left]

Working out the calculations, we get the final answers:

M=((3)+(1)2;(4)+(4)2)=(22;02)=(1;0)

We can compare this answer to the diagram above to check if the answer is reasonable: does the position of point M agree with the coordinates (1;0)? Indeed, that looks quite reasonable!

The coordinates of the mid-point are (1;0).


Submit your answer as:

Finding the mid-point from coordinates

There are two points, P(4;4) and N(103;4). What is the mid-point of P and N? Your answer should be exact (no rounding).

Your answer should be in brackets, with the ';' symbol between the coordinates. For example: ( 1/2 ; -5).

Answer: The mid-point is: .
coordinate
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Use the mid-point formula. Start by figuring out the values of x1,x2,y1 and y2.
STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 3 points left]

We have two points and we need to find their mid-point. For this we must use the mid-point formula:

M=(x1+x22;y1+y22)

In this equation, M represents the mid-point, which has the coordinates (xM;yM), while the variables x1, etc. refer to the coordinates of the end points. In this case, the end points are the points P and N. We must decide which point to use as point 1 and which to use as point 2. It does not matter which way we do it, so we will just use the order in which the coordinates were given in the question:

point 1:(4;4)point 2:(103;4)

This means:

x1=4x2=103y1=4y2=4

Plugging these values into the mid-point formula, we get:

(xM;yM)=((4)+(103)2;4+(4)2)

STEP: Substitute and evaluate each expression
[−2 points ⇒ 0 / 3 points left]

Now evaluate the expressions to get the answers. Remember that there are two different expressions in the mid-point formula, one for the x coordinates and the other for the y coordinates. It is helpful to separate the calculations, like this:

x coordinatesy coordinatesxM=(4)+(103)2yM=4+(4)2=(23)2=02=(23)×12=0=13

The resulting coordinates for the mid-point are (13;0). Notice that the y coordinate of the mid-point is zero, which happens because the y coordinates of P and N are opposite values.

We can do a simple check to see if this is reasonable by making a quick sketch of the points and the answer. You can see that the coordinates we got do in fact sit directly between points P and N.

The mid-point of P and N is at: (13;0).


Submit your answer as:

Finding the mid-point from coordinates

The points P and N have the coordinates (143;7) and (143;1), respectively. Compute the coordinates of the mid-point of P and N. Your answer should be exact (no rounding).

Your answer should be in brackets, with the ';' symbol between the coordinates. For example: ( 1/2 ; -5).

Answer: The mid-point is: .
coordinate
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Use the mid-point formula. Start by figuring out the values of x1,x2,y1 and y2.
STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 3 points left]

We have two points and we need to find their mid-point. For this we must use the mid-point formula:

M=(x1+x22;y1+y22)

In this equation, M represents the mid-point, which has the coordinates (xM;yM), while the variables x1, etc. refer to the coordinates of the end points. In this case, the end points are the points P and N. We must decide which point to use as point 1 and which to use as point 2. It does not matter which way we do it, so we will just use the order in which the coordinates were given in the question:

point 1:(143;7)point 2:(143;1)

This means:

x1=143x2=143y1=7y2=1

Plugging these values into the mid-point formula, we get:

(xM;yM)=((143)+(143)2;7+12)

STEP: Substitute and evaluate each expression
[−2 points ⇒ 0 / 3 points left]

Now evaluate the expressions to get the answers. Remember that there are two different expressions in the mid-point formula, one for the x coordinates and the other for the y coordinates. It is helpful to separate the calculations, like this:

x coordinatesy coordinatesxM=(143)+(143)2yM=7+12=02=82=0=4

The resulting coordinates for the mid-point are (0;4). Notice that the x coordinate of the mid-point is zero, which happens because the x coordinates of P and N are opposite values.

We can do a simple check to see if this is reasonable by making a quick sketch of the points and the answer. You can see that the coordinates we got do in fact sit directly between points P and N.

The mid-point of P and N is at: (0;4).


Submit your answer as:

Finding the mid-point from coordinates

There are two points, A(133;103) and B(133;3). What is the mid-point of A and B? Your answer should be exact (no rounding).

Your answer should be in brackets, with the ';' symbol between the coordinates. For example: ( 1/2 ; -5).

Answer: The mid-point is: .
coordinate
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
Use the mid-point formula. Start by figuring out the values of x1,x2,y1 and y2.
STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 3 points left]

We have two points and we need to find their mid-point. For this we must use the mid-point formula:

M=(x1+x22;y1+y22)

In this equation, M represents the mid-point, which has the coordinates (xM;yM), while the variables x1, etc. refer to the coordinates of the end points. In this case, the end points are the points A and B. We must decide which point to use as point 1 and which to use as point 2. It does not matter which way we do it, so we will just use the order in which the coordinates were given in the question:

point 1:(133;103)point 2:(133;3)

This means:

x1=133x2=133y1=103y2=3

Plugging these values into the mid-point formula, we get:

(xM;yM)=((133)+(133)2;(103)+32)

STEP: Substitute and evaluate each expression
[−2 points ⇒ 0 / 3 points left]

Now evaluate the expressions to get the answers. Remember that there are two different expressions in the mid-point formula, one for the x coordinates and the other for the y coordinates. It is helpful to separate the calculations, like this:

x coordinatesy coordinatesxM=(133)+(133)2yM=(103)+32=02=(13)2=0=(13)×12=16

The resulting coordinates for the mid-point are (0;16). Notice that the x coordinate of the mid-point is zero, which happens because the x coordinates of A and B are opposite values.

We can do a simple check to see if this is reasonable by making a quick sketch of the points and the answer. You can see that the coordinates we got do in fact sit directly between points A and B.

The mid-point of A and B is at: (0;16).


Submit your answer as:

Working with algebraic coordinates and the mid-point

What are the coordinates of a point B if the mid-point of B and another point A is at (p2;12) and the coordinates of A are (p2+2;3q)?

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of B is .
  2. The y-coordinate of B is .
expression
expression
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The question is about a mid-point: write down the mid-point formula and think about what values you can substitute into it.


STEP: Organise the information and make a plan
[−1 point ⇒ 3 / 4 points left]

In this question we have information about three points: we know the coordinates of one end point and the coordinates of the mid-points, but we do not know the coordinates of Point B.

Point A:(p2+2;3q)Point B:(xB;yB)the mid-point:(p2;12)

We do not know the coordinates of Point B, so we use variables to represent them.

The coordinates are given in terms of p and q, but do not let the variables distract you! The question is about a mid-point, and that means a point exactly halfway between two other points. In other words, the situation is something like the picture below, and we can solve it through the mid-point formula.

M=(x1+x22;y1+y22)

STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

Substitute in the values we have from above into the mid-point formula. The coordinates we have for the mid-point must go in on the left side.

(xM;yM)=(x1+x22;y1+y22)(p2;12)=((p2+2)+xB2;3q+yB2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The result above represents two separate equations: one for x and another for y. We can write them separately and solve each one. Begin by multiplying the equations by 2 to cancel the denominators.

x-coordinatesp2=(p2+2)+xB22(p2)=p2+2+xB2p2=p2+2+xB2p2p22=xBp22=xBy-coordinates12=3q+yB22(12)=3q+yB1=3q+yB13q=yB3q+1=yB

We cannot simplify the expressions any more, so we are done. The resulting answers are the coordinates for the end Point B: (p22;3q+1).

The final answers are:

  1. The x-coordinate of B is p22.
  2. The y-coordinate of B is 3q+1.

Submit your answer as: and

Working with algebraic coordinates and the mid-point

The mid-point of a Point Q(3x2;2y) and another Point R is at (3x2;32). What are the coordinates of Point R?

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of R is .
  2. The y-coordinate of R is .
expression
expression
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The question is about a mid-point: write down the mid-point formula and think about what values you can substitute into it.


STEP: Organise the information and make a plan
[−1 point ⇒ 3 / 4 points left]

In this question we have information about three points: we know the coordinates of one end point and the coordinates of the mid-points, but we do not know the coordinates of Point R.

Point Q:(3x2;2y)Point R:(xR;yR)the mid-point:(3x2;32)

We do not know the coordinates of Point R, so we use variables to represent them.

The coordinates are given in terms of x and y, but do not let the variables distract you! The question is about a mid-point, and that means a point exactly halfway between two other points. In other words, the situation is something like the picture below, and we can solve it through the mid-point formula.

M=(x1+x22;y1+y22)

STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

Substitute in the values we have from above into the mid-point formula. The coordinates we have for the mid-point must go in on the left side.

(xM;yM)=(x1+x22;y1+y22)(3x2;32)=((3x2)+xR2;2y+yR2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The result above represents two separate equations: one for x and another for y. We can write them separately and solve each one. Begin by multiplying the equations by 2 to cancel the denominators.

x-coordinates3x2=(3x2)+xR22(3x2)=3x2+xR6x2=3x2+xR6x2+3x2=xR3x2=xRy-coordinates32=2y+yR22(32)=2y+yR3=2y+yR32y=yR2y+3=yR

We cannot simplify the expressions any more, so we are done. The resulting answers are the coordinates for the end Point R: (3x2;2y+3).

The final answers are:

  1. The x-coordinate of R is 3x2.
  2. The y-coordinate of R is 2y+3.

Submit your answer as: and

Working with algebraic coordinates and the mid-point

What are the coordinates of a point N if the mid-point of N and another point P is at (72;0) and the coordinates of P are (3x+3;y2+4)?

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of N is .
  2. The y-coordinate of N is .
expression
expression
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The question is about a mid-point: write down the mid-point formula and think about what values you can substitute into it.


STEP: Organise the information and make a plan
[−1 point ⇒ 3 / 4 points left]

In this question we have information about three points: we know the coordinates of one end point and the coordinates of the mid-points, but we do not know the coordinates of Point N.

Point P:(3x+3;y2+4)Point N:(xN;yN)the mid-point:(72;0)

We do not know the coordinates of Point N, so we use variables to represent them.

The coordinates are given in terms of x and y, but do not let the variables distract you! The question is about a mid-point, and that means a point exactly halfway between two other points. In other words, the situation is something like the picture below, and we can solve it through the mid-point formula.

M=(x1+x22;y1+y22)

STEP: Substitute the values into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

Substitute in the values we have from above into the mid-point formula. The coordinates we have for the mid-point must go in on the left side.

(xM;yM)=(x1+x22;y1+y22)(72;0)=((3x+3)+xN2;(y2+4)+yN2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The result above represents two separate equations: one for x and another for y. We can write them separately and solve each one. Begin by multiplying the equations by 2 to cancel the denominators.

x-coordinates72=(3x+3)+xN22(72)=3x+3+xN7=3x+3+xN73x3=xN3x+4=xNy-coordinates0=(y2+4)+yN22(0)=y2+4+yN0=y2+4+yN0y24=yNy24=yN

We cannot simplify the expressions any more, so we are done. The resulting answers are the coordinates for the end Point N: (3x+4;y24).

The final answers are:

  1. The x-coordinate of N is 3x+4.
  2. The y-coordinate of N is y24.

Submit your answer as: and

Using a mid-point to find an end point

As shown on the picture below, the mid-point of A and another Point B is at (174;14) and the coordinates of B are (52;52).

What are the coordinates of a Point A?

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of A is .
  2. The y-coordinate of A is .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the question includes a mid-point, think about how you can use the mid-point formula.


STEP: Set up the equations based on the mid-point formula
[−2 points ⇒ 2 / 4 points left]

This question is about a mid-point, which as always is a point exactly between two other points. In this case, however, we only know the coordinates of one of those end points:

Point A:(xA;yA)Point B:(52;52)the mid-point:(174;14)

We do not know the coordinates of Point A, so we use variables to represent them.

Since this question is about a mid-point, and we know the coordinates of some of the points involved, we should approach the question through the mid-point formula:

M=(x1+x22;y1+y22)(xM;yM)=(x1+x22;y1+y22)

In this case, we know that the mid-point coordinates are (174;14), but we do not know the coordinates of Point A. So the equation becomes:

(174;14)=(xA+(52)2;yA+(52)2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The equation above actually represents two separate equations: one for x and another for y. Let's write those two equations separately, and solve! Start by cancelling the denominators in the equations. We do this by multiplying both sides of the equation by the LCD of the fractions in the equation. In this case, the LCD matches the denominator on the left side of the equation for both equations: we will multiply both equations by 4. (Notice that when the 4 cancels the denominator of 2, it leaves a factor of 2 which we must distribute into the binomial.)

x-coordinates174=xA+(52)24(174)=(xA+(52)2)417=(xA52)217=2xA517+5=2xA12=2xA6=xAy-coordinates14=yA+(52)24(14)=(yA+(52)2)41=(yA52)21=2yA51+5=2yA6=2yA3=yA

We get the answers xA = 6 and yA = 3. Compare these numbers to the graph in the question: we can see that the answer is consistent with the position of Point A on the graph.

Point A must be at (6;3), which means the final answers are:

  1. The x-coordinate of A is −6.
  2. The y-coordinate of A is 3.

Submit your answer as: and

Using a mid-point to find an end point

As shown on the picture below, the mid-point of A and another Point B is at (54;134) and the coordinates of B are (32;32).

What are the coordinates of a Point A?

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of A is .
  2. The y-coordinate of A is .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the question includes a mid-point, think about how you can use the mid-point formula.


STEP: Set up the equations based on the mid-point formula
[−2 points ⇒ 2 / 4 points left]

This question is about a mid-point, which as always is a point exactly between two other points. In this case, however, we only know the coordinates of one of those end points:

Point A:(xA;yA)Point B:(32;32)the mid-point:(54;134)

We do not know the coordinates of Point A, so we use variables to represent them.

Since this question is about a mid-point, and we know the coordinates of some of the points involved, we should approach the question through the mid-point formula:

M=(x1+x22;y1+y22)(xM;yM)=(x1+x22;y1+y22)

In this case, we know that the mid-point coordinates are (54;134), but we do not know the coordinates of Point A. So the equation becomes:

(54;134)=(xA+(32)2;yA+(32)2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The equation above actually represents two separate equations: one for x and another for y. Let's write those two equations separately, and solve! Start by cancelling the denominators in the equations. We do this by multiplying both sides of the equation by the LCD of the fractions in the equation. In this case, the LCD matches the denominator on the left side of the equation for both equations: we will multiply both equations by 4. (Notice that when the 4 cancels the denominator of 2, it leaves a factor of 2 which we must distribute into the binomial.)

x-coordinates54=xA+(32)24(54)=(xA+(32)2)45=(xA32)25=2xA35+3=2xA8=2xA4=xAy-coordinates134=yA+(32)24(134)=(yA+(32)2)413=(yA32)213=2yA313+3=2yA10=2yA5=yA

We get the answers xA = 4 and yA = 5. Compare these numbers to the graph in the question: we can see that the answer is consistent with the position of Point A on the graph.

Point A must be at (4;5), which means the final answers are:

  1. The x-coordinate of A is 4.
  2. The y-coordinate of A is −5.

Submit your answer as: and

Using a mid-point to find an end point

The mid-point of Point A(5;7) and another Point B is at (3;234). This is represented on the diagram below.

What are the coordinates of Point B?

INSTRUCTION: Give your answers as two separate coordinates, as indicated below.
Answer:
  1. The x-coordinate of B is .
  2. The y-coordinate of B is .
numeric
numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the question includes a mid-point, think about how you can use the mid-point formula.


STEP: Set up the equations based on the mid-point formula
[−2 points ⇒ 2 / 4 points left]

This question is about a mid-point, which as always is a point exactly between two other points. In this case, however, we only know the coordinates of one of those end points:

Point A:(5;7)Point B:(xB;yB)the mid-point:(3;234)

We do not know the coordinates of Point B, so we use variables to represent them.

Since this question is about a mid-point, and we know the coordinates of some of the points involved, we should approach the question through the mid-point formula:

M=(x1+x22;y1+y22)(xM;yM)=(x1+x22;y1+y22)

In this case, we know that the mid-point coordinates are (3;234), but we do not know the coordinates of Point B. So the equation becomes:

(3;234)=((5)+xB2;7+yB2)

STEP: Solve the equations for the coordinate values
[−2 points ⇒ 0 / 4 points left]

The equation above actually represents two separate equations: one for x and another for y. Let's write those two equations separately, and solve! Start by cancelling the denominators in the equations. We do this by multiplying both sides of the equation by the LCD of the fractions in the equation. We will multiply the x equation by 2, while we will multiply the y equation by 4. (Notice that when the 4 cancels the denominator of 2, it leaves a factor of 2 which we must distribute into the binomial.)

x-coordinates3=(5)+xB22(3)=((5)+xB2)26=5+xB6+5=xB1=xBy-coordinates234=7+yB24(234)=(7+yB2)423=(7+yB)223=14+2yB2314=2yB9=2yB92=yB

We get the answers xB = 1 and yB = 92. Compare these numbers to the graph in the question: we can see that the answer is consistent with the position of Point B on the graph.

Point B must be at (1;92), which means the final answers are:

  1. The x-coordinate of B is −1.
  2. The y-coordinate of B is 92.

Submit your answer as: and

Mid-point calculations with algebra

Find the mid-point of the points here:

C(3p2+5;q)D(3p2+2;q),p,qN0

Type your answer with brackets and the ';' symbol between the coordinates. For example, your answer might look like this: ( a + 3 ; -2b ).

Answer:
The mid-point coordinates are: .
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Use the mid-point formula. If the variables seem to make the question complex, try to ignore them and simply think about working with the formula.
STEP: Organise the information in the question
[−1 point ⇒ 3 / 4 points left]

This question is about finding a mid-point. It is complex because the coordinates contain variables, but in the end it is still a mid-point calculation. Start by writing down the information from the question. (Notice that the information "p,qN0" does not matter: for the mid-point it is the coordinate values which we need.)

point C:(3p2+5;q)point D:(3p2+2;q)the mid-point:(xM;yM)coordinates so we must use variablesWe do not know the mid-point

Remember that this is about a mid-point, which means the situation looks somehow like the picture below. We have no idea where the points actually are, but the situation is something like the picture.


STEP: Substitute into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

We need to substitute the coordinates above into the mid-point formula because we want to find the mid-point of points C and D. The mid-point formula is:

M=(x1+x22;y1+y22)

where x1,y1, etc. represent the coordinates of the end points (C and D). So:

x1=3p2+5,x2=3p2+2,y1=q,y2=q

Substitute these expressions into the formula. Make sure you use brackets around the substitutions to help stay organised.

(xM;yM)=((3p2+5)+(3p2+2)2;(q)+(q)2)

STEP: Simplify each expression
[−2 points ⇒ 0 / 4 points left]

Now we need to simplify as much as possible. Remember that there are two different expressions to evaluate in the mid-point formula, one for x and one for y. We will separate these calculations to simplify things.

x coordinatesy coordinatesxM=(3p2+5)+(3p2+2)2yM=(q)+(q)2=3p2+5+3p2+22=qq2=6p2+72=2q2=6p22+72=q=3p2+72

There we have the answer! The mid-point of the two points is: (3p2+72;q).


Submit your answer as:

Mid-point calculations with algebra

Calculate the mid-point of the points here:

M(2h+2;3j25)P(2h;3j2),h,jN0

Type your answer with brackets and the ';' symbol between the coordinates. For example, your answer might look like this: ( a + 3 ; -2b ).

Answer:
The mid-point coordinates are: .
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Use the mid-point formula. If the variables seem to make the question complex, try to ignore them and simply think about working with the formula.
STEP: Organise the information in the question
[−1 point ⇒ 3 / 4 points left]

This question is about finding a mid-point. It is complex because the coordinates contain variables, but in the end it is still a mid-point calculation. Start by writing down the information from the question. (Notice that the information "h,jN0" does not matter: for the mid-point it is the coordinate values which we need.)

point M:(2h+2;3j25)point P:(2h;3j2)the mid-point:(xM;yM)coordinates so we must use variablesWe do not know the mid-point

Remember that this is about a mid-point, which means the situation looks somehow like the picture below. We have no idea where the points actually are, but the situation is something like the picture.


STEP: Substitute into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

We need to substitute the coordinates above into the mid-point formula because we want to find the mid-point of points M and P. The mid-point formula is:

M=(x1+x22;y1+y22)

where x1,y1, etc. represent the coordinates of the end points (M and P). So:

x1=2h+2,x2=2h,y1=3j25,y2=3j2

Substitute these expressions into the formula. Make sure you use brackets around the substitutions to help stay organised.

(xM;yM)=((2h+2)+(2h)2;(3j25)+(3j2)2)

STEP: Simplify each expression
[−2 points ⇒ 0 / 4 points left]

Now we need to simplify as much as possible. Remember that there are two different expressions to evaluate in the mid-point formula, one for x and one for y. We will separate these calculations to simplify things.

x coordinatesy coordinatesxM=(2h+2)+(2h)2yM=(3j25)+(3j2)2=2h+2+2h2=3j25+3j22=22=6j252=1=6j2252=3j252

There we have the answer! The mid-point of the two points is: (1;3j252).


Submit your answer as:

Mid-point calculations with algebra

The points C and D have the following coordinates:

C(3x2;3y2)D(3x2;3y2+4),x,yZ

Calculate the coordinates of the mid-point of C and D. Type your answer with brackets and the ';' symbol between the coordinates. For example, your answer might look like this: ( a + 3 ; -2b ).

Answer:
The mid-point coordinates are: .
coordinate
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]
Use the mid-point formula. If the variables seem to make the question complex, try to ignore them and simply think about working with the formula.
STEP: Organise the information in the question
[−1 point ⇒ 3 / 4 points left]

This question is about finding a mid-point. It is complex because the coordinates contain variables, but in the end it is still a mid-point calculation. Start by writing down the information from the question. (Notice that the information "x,yZ" does not matter: for the mid-point it is the coordinate values which we need.)

point C:(3x2;3y2)point D:(3x2;3y2+4)the mid-point:(xM;yM)coordinates so we must use variablesWe do not know the mid-point

Remember that this is about a mid-point, which means the situation looks somehow like the picture below. We have no idea where the points actually are, but the situation is something like the picture.


STEP: Substitute into the mid-point formula
[−1 point ⇒ 2 / 4 points left]

We need to substitute the coordinates above into the mid-point formula because we want to find the mid-point of points C and D. The mid-point formula is:

M=(x1+x22;y1+y22)

where x1,y1, etc. represent the coordinates of the end points (C and D). So:

x1=3x2,x2=3x2,y1=3y2,y2=3y2+4

Substitute these expressions into the formula. Make sure you use brackets around the substitutions to help stay organised.

(xM;yM)=((3x2)+(3x2)2;(3y2)+(3y2+4)2)

STEP: Simplify each expression
[−2 points ⇒ 0 / 4 points left]

Now we need to simplify as much as possible. Remember that there are two different expressions to evaluate in the mid-point formula, one for x and one for y. We will separate these calculations to simplify things.

x coordinatesy coordinatesxM=(3x2)+(3x2)2yM=(3y2)+(3y2+4)2=3x23x22=3y2+3y2+42=02=6y2+42=0=2(3y2+2)2=3y2+2

There we have the answer! The mid-point of the two points is: (0;3y2+2).


Submit your answer as:

5. Angles

Finding the equation of a specific line.

Determine the equation of a line passing through the point (2;5) which is parallel to the line

y=5x3+4

INSTRUCTION: Type the entire equation in standard form, which will either be y=mx+c or x=c.
Answer: The equation is: .
equation
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line! Start by finding the gradient of the line you want.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line we want. From the given equation we can read off the gradient: m=53.

For a line which is parallel to this, we must use the same gradient: the gradient of the line we want is also m=53.


STEP: Find the y-intercept of the line
[−1 point ⇒ 1 / 3 points left]

Now substitute the coordinates (2;5), together with the gradient, into the equation to find the value of the y-intercept, c.

y=mx+c(5)=(53)(2)+c5=103+c53=c

STEP: Write the final equation
[−1 point ⇒ 0 / 3 points left]

Now we just put all of the pieces together into standard form. The equation that we want is

y=5x353

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (2;5) is shown with a black dot. We can see that the red line passes through the point just like it is supposed to! It is also clear that the two lines are parallel.

The equation of the line is y=5x353.


Submit your answer as:

Finding equations of perpendicular lines

The graph here shows a line segment, AB¯. The blue dashed line is perpendicular to AB¯.

The equation of the blue dashed line is y=2x+0,75. Point A is at (2;0).

Determine the equation of the line segment AB¯.

INSTRUCTION: Type only the right hand side of the equation into the box below.
Answer: y =
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The gradient of the line AB is equal to: 1gradientPL, where PL stands for Perpendicular Line.


STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of AB¯. We can get the gradient from the blue dashed line.

The gradients of perpendicular lines are negative reciprocals of each other. Therefore the gradient of AB¯ must be equal to: 1gradientPL, where PL stands for Perpendicular Line. From the equation given for the blue dashed line, we know that its gradient is −2.

y=mx+cy=(1gradientPL)x+cy=(12)x+cy=(0,5)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(0)=(0,5)(2)+cc=1

STEP: Write the equation of the line in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=0,5x+1

Submit your answer as:

Calculating the inclination of a line from two points

Determine the inclination of the line (θ). Round your answer to 2 decimal places.

Answer: θ = °
numeric
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

To calculate the inclination of a straight line, we first need to calculate the gradient:

m=yByAxBxA=(0,5)(3)(5)(2)m=0,5

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

Now that we know the gradient, we use the following equation to determine the value of θ:

m=tan(θ)θ=tan1(m)=tan1(0,5)=26,57°

Very important: always check that your calculator is in DEG mode when using the trigonometric function buttons.


STEP: <no title>
[−0 points ⇒ 0 / 2 points left]

If the gradient of the line is negative, we add 180° to the above angle to get an obtuse angle. In this case, the gradient is positive.

θ=26,57°

Submit your answer as:

Parallel and perpendicular lines

Determine whether the equations here represent lines which are parallel, perpendicular, or neither.

2x+2y+5=0x+2y=x+5

Answer choices:

  • parallel
  • perpendicular
  • neither
INSTRUCTION: Type your answer from the three choices given; mispelled answers will be marked wrong.
Answer: The lines are .
string
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To answer this question, you must know the gradients of each equation. To find the gradients, start by arranging the equations in standard form for linear equations:

y=mx+c

STEP: Arrange the first equation in standard form
[−1 point ⇒ 2 / 3 points left]

We have two equations and need to determine if they represent lines which are parallel, perpendicular or neither.

We can only answer this question by comparing the gradients of the two equations. So we need to arrange the equations in standard form. For the first equation this will be:

2x+2y+5=02y=2x512(2y)=(2x5)12y=x52

The gradient of the first equation is m1=1.


STEP: Arrange the second equation in standard form
[−1 point ⇒ 1 / 3 points left]

Now change the second equation into standard form.

x+2y=x+52y=2x+512(2y)=(2x+5)12y=x+52

The gradient of the second equation is m2=1.


STEP: Compare the gradients of the equations
[−1 point ⇒ 0 / 3 points left]

Now we can compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those. The key facts to remember are that:

  • if the gradients are equal the lines are parallel
  • if the gradients are negative reciprocals the lines are perpendicular

The equations have equal gradients, so they must be parallel lines.

The two equations are graphed on the Cartesian plane below. We can see that the two lines run parallel to one another.

The correct answer is parallel.


Submit your answer as:

Calculating equations of perpendicular lines

Consider the following diagram:

Line AB is perpendicular to the line y=2x2,25 (blue line).

Calculate the equation of AB.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of a straight line equation is y=mx+c.

To calculate the equation of the straight line, we first need the gradient (m) of AB, which we can get from the perpendicular line:

For perpendicular lines, we know that m1×m2=1:

mAB×2=1mAB=12=0,5

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we substitute the coordinates of A into the equation to calculate the value of the y-intercept (c):

y=0,5x+c3=(0,5)(2)+cc=2

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is:

y=0,5x2

Submit your answer as:

Using inclination to find the equation of a line

The line shown below has an angle of inclination θ = 153,43°.

Find the equation of the line. Round your answer to one decimal place.

Answer:y =
expression
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

We are given the angle of inclination of the line, which we can use to calculate the gradient:

m=tanθ=tan(153,43°)=0,5

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

We know the general form of the equation of a straight line is y=mx+c.

Substitute the value for m into the equation:

y=0,5x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

Next we calculate the value of c by substituting the coordinates of point A(3;3.5) into the equation and simplifying:

y=0,5x+c3,5=(0,5)(3)+cc=2

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Therefore, the equation of the line AB is

y=0,5x+2

Submit your answer as:

Finding the equation of a line

What is the equation of the line which is perpendicular to y=6x2 and passes through (1;5)?

Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.

Answer:
equation
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. From the given equation you can see that the gradient is m=6.

For lines that are perpendicular, the gradient must be the negative reciprocal, therefore m=16.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Substitute the coordinates (1;5) and the gradient into the equation to find the value of c.

y=mx+c5=(16)(1)+c5=16+c316=c

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is y=x6316.

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (1;5) is shown with a black dot labelled with an A. You can see that the red line passes through the point and that the lines are perpendicular to each other.


Submit your answer as:

Finding a coordinate of a point on a perpendicular line

You are given the following diagram, which shows two perpendicular lines on the Cartesian plane. Line AB has endpoints at A(3;1) and B(2;y). The other line (the dashed line) has the equation y=x+32.

Calculate the missing value of y.

Answer:

y=

numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The gradient of the line AB is equal to:

1gradientPL

where PL stands for Perpendicular Line.


STEP: Write down the equation of a straight line
[−1 point ⇒ 3 / 4 points left]

We need to find the equation of line AB. So we will need the equation of a straight line:

y=mx+c

STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 4 points left]

The line AB has a gradient that is equal to:

1gradientPL

where PL stands for Perpendicular Line. The gradient of the perpendicular line is 1, so:

mAB=1gradientPL=1(1)=1

Now we know that the equation of line AB is

y=x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known point A into the equation, to find c:

y=x+c(1)=(3)+cc=2

And now we have the complete equation for line AB:

y=x+2

STEP: Substitute the known value for point B into the equation and solve
[−1 point ⇒ 0 / 4 points left]

We can use the equation for line AB to find the missing coordinate. To do this, substitute the known value for point B into the equation for line AB:

y=x+2y=(2)+2

Solving, we get:

y=4

The correct answer is y=4.


Submit your answer as:

Parallel and perpendicular lines

Determine whether the equations represent lines which are parallel, perpendicular, or neither.

6x=2y32y+3=6x+8
Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, we need to know the gradient of each equation. This means that we need to start by arranging the equations in standard form for linear equations: y=mx+c.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

We answer this question by comparing the gradients of the two equations, for which the equations have to be in standard form. For the first equation this is:

6x=2y32y=6x312(2y)=12(6x3)y=3x32

The gradient of the first equation is m=3.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now we change the second equation into standard form.

2y+3=6x+82y=6x+512(2y)=12(6x+5)y=3x+52

The gradient of the second equation is m=3.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

We now compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those.

  • Are the gradients equal to each other (parallel gradients)?
  • Do they have a product of 1 (perpendicular gradients)?
  • Are neither of these true?

The equations have equal gradients, so they must be parallel lines. The correct response is: parallel.

The two equations are graphed on the Cartesian plane shown below. We can see that the two lines run parallel to each another.


Submit your answer as:

Calculating inclination from two points

Consider the diagram below, which shows a line through the points: A(4;132) and B(5;7).

Calculate the inclination of the line, which is the angle θ. Round your answer to 2 decimal places if appropriate.

Answer: θ = °
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need the inclination of the line, which is the angle θ. The inclination is related to the gradient, so start by finding the gradient of the line.


STEP: Find the gradient of the line
[−1 point ⇒ 1 / 2 points left]

We need to find the inclination of the line. The inclination is the angle between the line and the positive x-axis. This is shown by the angle θ in the diagram.

The inclination of a line is directly related to the line's gradient: both represent the steepness of the line. (For an explanation about the connection between the gradient and the inclination, you can read here.) So to find the inclination, we need the gradient of the line.

m=y2y1x2x1=(7)(132)(5)(4)=32

STEP: Calculate the inclination using the inclination formula
[−1 point ⇒ 0 / 2 points left]

Now we can calculate the inclination of the line. We do this using the following equation:

m=tanθθ=tan1(m)=tan1(32)=56,30993...°56,31°

STEP: Adjust the answer if necessary
[−0 points ⇒ 0 / 2 points left]

If the gradient is negative we must add 180° to the above value to get an obtuse angle. In this case, the gradient, and the angle we got above, is positive.

θ56,31°

The inclination of the line is 56,31°.


Submit your answer as:

Working with perpendicular lines

In the diagram below, the two lines are perpendicular. The equation of the dashed red line is

y=x272

As labelled below, the solid blue line passes through the points A(x;12) and B(2;112).

Determine xA, the x-coordinate of point A.

Answer: xA =
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by finding the gradient of line AB. You can do this because the lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 4 / 5 points left]

We need to find the equation of the blue line, which is called line AB. That means we need the gradient and the y-intercept of line AB.

Since the two lines in the figure are perpendicular to each other, we can find the gradient of line AB from the gradient of the dashed red line. Perpendicular gradients are negative reciprocals of each other.

The gradient of the dashed red line is 12 (we can read it straight from the equation), so:

mAB=negative reciprocal of mred line=reciprocal of 12=(2)=2

STEP: Start building the equation
[−1 point ⇒ 3 / 5 points left]

Since the equation for a staight line is:

y=mx+c

The equation for line AB must be:

y=2x+c

STEP: Find the y-intercept of the equation
[−1 point ⇒ 2 / 5 points left]

Now we substitute the coordinates of point B into the equation, to find c:

y=2x+c112=2(2)+c32=c

Now we know the equation for line AB:

y=2x32

STEP: Use the y-coordinate of point A to find xA
[−2 points ⇒ 0 / 5 points left]

Now we use a key concept for graphs and their equations: every point on the line agrees with the equation. If we substitute in the y-coordinate of any point on the line into the equation, we will always get the x-coordinate for that point.The equation forces this to be true! So now substitute the y-coordinate for point A into the equation for line AB and solve for xA.

yA=2xA32(12)=2xA32xA=1

The correct answer is xA=1.


Submit your answer as:

Facts about perpendicular lines

  1. From the following table, which statement or statements are true about perpendicular lines and the gradients of perpendicular lines? Identify all of the true statements.

    A m1×m2=1
    B Perpendicular gradients are negative reciprocals of each other.
    C Perpendicular gradients have a product of 1.
    D Perpendicular gradients are equal.
    INSTRUCTION: There may be more than one true statement above. You need to identify all true statements.
    Answer: Statement(s) is (are) true
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    There is only one true statement in the table above.
    STEP: Use facts about perpendicular lines
    [−1 point ⇒ 0 / 1 points left]

    The following statements are true for the gradients of perpendicular lines:

    • The gradients must be negative reciprocals of each other. For example, the negative reciprocal of 1 is 1.
    • The gradients must have a product of -1 : m1×m2=1. For example, 1 and 1 are perpendicular gradients because
      (1)×(1)=1
      .

    There is only one statement above which agrees with these facts.

    The correct choice is: B.


    Submit your answer as:
  2. If a certain line has a gradient of m=32, what is the gradient of a line perpendicular to that line? (Remember that division by zero is 'undefined.')

    Answer: m=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the facts from Question 1 to work out the answer!


    STEP: Find the negative reciprocal of the given gradient
    [−1 point ⇒ 0 / 1 points left]

    If the gradients are perpendicular, then the gradient we want must be the negative reciprocal of the gradient given. 'Negative' means change the sign (multiply by -1) and 'reciprocal' means turn the number upside down. The gradient given is m=32: the negative of that is 32. Then the reciprocal of that is 23.

    Note: for perpendicular gradients, m1×m2=1. The gradient given is m=32, so we want to solve (32)×m2=1. The value which will solve that equation is 23.

    Finally, we can look at a graph of lines which have the gradients above. You can see that they are perpendicular to each other! You can also see that one of the lines has a positive gradient while the other one has a negative gradient - they are negatives.

    The correct answer is 23.


    Submit your answer as:

Finding the equation of a perpendicular line

The figure below shows two perpendicular lines. The equation of the dashed red line is:

y=3x+52

The solid blue line passes through points A and B. The coordinates of point A are (3;72) and the coordinates of B are unknown.

Calculate the equation of line AB.

INSTRUCTION: Type only the right side of the equation into the answer box (following the "y = ...").
Answer: Equation of line AB: y =
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the gradient of line AB. You can do this using the fact that the two lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

The equation of a straight line requires the gradient and the y-intercept for the line. We can find the gradient in this case because line AB is perpendicular to the dashed red line: if we know a gradient, we can always find the gradient which is perpendicular to it.

The equation of the dashed red line is

y=3x+52

which means that the gradient of the dashed line is 3.

Perpendicular lines have gradients which are negative reciprocals of each other. So the gradient of line AB is the negative reciprocal of 3.

mAB=negative reciprocal of mdashed line=reciprocal of 3=(13)=13

This means that the equation of line AB is:

y=x3+c

STEP: Find the value of c
[−1 point ⇒ 1 / 3 points left]

Now we can find the value of c for the line (which is the y-intercept). We do this by substituting any point on the line into the equation from above. We only know the coordinates of point A, which are (3;72). Substitute these coordinates into the equation and solve for c.

y=x3+c(72)=(3)3+ccoordinates of point ASubstitute in the52=c

STEP: Write the equation
[−1 point ⇒ 0 / 3 points left]

We now have both numbers we need for the equation of line AB: we just need to put them together.

The correct answer is:

y=x352

Submit your answer as:

Finding a point using perpendicular lines

The following diagram is given:

Line AB is perpendicular to the line y=2x2,25.

Determine the x-coordinate of point B, rounded to 2 decimal places.

Answer: B (;3)
numeric
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Recall the standard form of a straight line equation:

y=mx+c

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

For perpendicular lines, we know that m1×m2=1:

mAB×2=1mAB=12=0,5y=0,5x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

We substitute the x and y coordinates of point A into the equation to find c:

y=0,5x+c1,5=(0,5)(1)+cc=2

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Finally, we substitute the y-coordinate for point B into the equation and solve for x:

y=0,5x23=(0,5)(x)2x=2

Submit your answer as:

Finding the equation of a line from the inclination

The diagram below shows a line on the Cartesian plane. The line passes through the point A(6;72). The line has an angle of inclination θ=135°.

Find the equation of the line.

INSTRUCTIONS:
  • Give only the right side of the equation, following y=....
  • Round the values in the equation to 2 decimal places if necessary.
Answer: Complete the equation for the line: y = .
one-of
type(expression)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the inclination to find the gradient of the line. Then use that gradient to start working out the equation.


STEP: Use the inclination to find the gradient of the line
[−1 point ⇒ 3 / 4 points left]

This question is about finding the equation of a straight line. We know one point on the line and the line's inclination. The equation of the line will have the form y=mx+c. So we need to find the gradient, m, and the y-intercept, c.

The inclination of a line tells us about the steepness of the line so it is similar to the gradient. In fact, inclination and gradient are directly related to each other. We can use the following equation to calculate the gradient from the inclination.

m=tanθ=tan(135°)=1

The gradient is 1. Note that the gradient is negative, which agrees with the slope of the graph.


STEP: Put the gradient into the equation
[−1 point ⇒ 2 / 4 points left]

Now we recall the general form of the equation of a straight line. Since we know the gradient m of the line, we can substitute that into the equation of the line:

y=(1)x+c

STEP: Use the point to find the y-intercept
[−1 point ⇒ 1 / 4 points left]

Now we can calculate the value of the y-intercept of the line (the value of c). We do this by substituting the coordinates of any point on the line into the general form for the line from above. Since we know the coordinates of point A, we will use those coordinates, which are A(6;72).

y=x+c(72)=(6)+c2,5=c

Remember that this number should be the y-intercept of the graph. As with the gradient, we can compare this answer to the graph to check if they agree: does the graph up above have a y-intercept close to 2,5? Yes!


STEP: Write the complete equation
[−1 point ⇒ 0 / 4 points left]

We now have the gradient and the y-intercept of the line. The equation of the line is:

y=x+2,5

Submit your answer as:

Exercises

Finding the equation of a specific line.

What is the equation of a line which is perpendicular to the line

y=2x3+1

and passes through (1;5)?

INSTRUCTION: Type the entire equation in standard form, which will either be y=mx+c or x=c.
Answer: The equation is: .
equation
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line! Start by finding the gradient of the line you want.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line we want. From the given equation we can read off the gradient: m=23.

For a line which is perpendicular to this, the gradient must be the negative reciprocal: m=32.


STEP: Find the y-intercept of the line
[−1 point ⇒ 1 / 3 points left]

Now substitute the coordinates (1;5), together with the gradient, into the equation to find the value of the y-intercept, c.

y=mx+c(5)=(32)(1)+c5=32+c132=c

STEP: Write the final equation
[−1 point ⇒ 0 / 3 points left]

Now we just put all of the pieces together into standard form. The equation that we want is

y=3x2+132

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (1;5) is shown with a black dot. We can see that the red line passes through the point just like it is supposed to! It is also clear that the lines are perpendicular to each other.

The equation of the line is y=3x2+132.


Submit your answer as:

Finding the equation of a specific line.

Determine the equation of a line passing through the point (3;1) which is perpendicular to the line

y=5x2+4

INSTRUCTION: Type the entire equation in standard form, which will either be y=mx+c or x=c.
Answer: The equation is: .
equation
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line! Start by finding the gradient of the line you want.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line we want. From the given equation we can read off the gradient: m=52.

For a line which is perpendicular to this, the gradient must be the negative reciprocal: m=25.


STEP: Find the y-intercept of the line
[−1 point ⇒ 1 / 3 points left]

Now substitute the coordinates (3;1), together with the gradient, into the equation to find the value of the y-intercept, c.

y=mx+c(1)=(25)(3)+c1=65+c115=c

STEP: Write the final equation
[−1 point ⇒ 0 / 3 points left]

Now we just put all of the pieces together into standard form. The equation that we want is

y=2x5+115

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (3;1) is shown with a black dot. We can see that the red line passes through the point just like it is supposed to! It is also clear that the lines are perpendicular to each other.

The equation of the line is y=2x5+115.


Submit your answer as:

Finding the equation of a specific line.

What is the equation of a line which is parallel to the line

y=6x5+4

and passes through (1;3)?

INSTRUCTION: Type the entire equation in standard form, which will either be y=mx+c or x=c.
Answer: The equation is: .
equation
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line! Start by finding the gradient of the line you want.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line we want. From the given equation we can read off the gradient: m=65.

For a line which is parallel to this, we must use the same gradient: the gradient of the line we want is also m=65.


STEP: Find the y-intercept of the line
[−1 point ⇒ 1 / 3 points left]

Now substitute the coordinates (1;3), together with the gradient, into the equation to find the value of the y-intercept, c.

y=mx+c(3)=(65)(1)+c3=65+c215=c

STEP: Write the final equation
[−1 point ⇒ 0 / 3 points left]

Now we just put all of the pieces together into standard form. The equation that we want is

y=6x5215

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (1;3) is shown with a black dot. We can see that the red line passes through the point just like it is supposed to! It is also clear that the two lines are parallel.

The equation of the line is y=6x5215.


Submit your answer as:

Finding equations of perpendicular lines

The graph here shows a line segment, AB¯. The blue dashed line is perpendicular to AB¯.

The equation of the blue dashed line is y=2x0,75. Point A is at (5;1,5).

Determine the equation of the line segment AB¯.

INSTRUCTION: Type only the right hand side of the equation into the box below.
Answer: y =
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The gradient of the line AB is equal to: 1gradientPL, where PL stands for Perpendicular Line.


STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of AB¯. We can get the gradient from the blue dashed line.

The gradients of perpendicular lines are negative reciprocals of each other. Therefore the gradient of AB¯ must be equal to: 1gradientPL, where PL stands for Perpendicular Line. From the equation given for the blue dashed line, we know that its gradient is 2.

y=mx+cy=(1gradientPL)x+cy=(12)x+cy=(0,5)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(1,5)=(0,5)(5)+cc=1

STEP: Write the equation of the line in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=0,5x1

Submit your answer as:

Finding equations of perpendicular lines

The graph here shows a line segment, AB¯. The blue dashed line is perpendicular to AB¯.

The equation of the blue dashed line is y=2x0,75. Point A is at (4;3).

Determine the equation of the line segment AB¯.

INSTRUCTION: Type only the right hand side of the equation into the box below.
Answer: y =
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The gradient of the line AB is equal to: 1gradientPL, where PL stands for Perpendicular Line.


STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of AB¯. We can get the gradient from the blue dashed line.

The gradients of perpendicular lines are negative reciprocals of each other. Therefore the gradient of AB¯ must be equal to: 1gradientPL, where PL stands for Perpendicular Line. From the equation given for the blue dashed line, we know that its gradient is −2.

y=mx+cy=(1gradientPL)x+cy=(12)x+cy=(0,5)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(3)=(0,5)(4)+cc=1

STEP: Write the equation of the line in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=0,5x1

Submit your answer as:

Finding equations of perpendicular lines

The graph here shows a line segment, AB¯. The blue dashed line is perpendicular to AB¯.

The equation of the blue dashed line is y=2x0,75. Point A is at (5;1).

Determine the equation of the line segment AB¯.

INSTRUCTION: Type only the right hand side of the equation into the box below.
Answer: y =
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The gradient of the line AB is equal to: 1gradientPL, where PL stands for Perpendicular Line.


STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 3 points left]

To calculate the equation of the straight line, we first need the gradient (m) of AB¯. We can get the gradient from the blue dashed line.

The gradients of perpendicular lines are negative reciprocals of each other. Therefore the gradient of AB¯ must be equal to: 1gradientPL, where PL stands for Perpendicular Line. From the equation given for the blue dashed line, we know that its gradient is 2.

y=mx+cy=(1gradientPL)x+cy=(12)x+cy=(0,5)x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 3 points left]

Secondly, we calculate the value of the y-intercept (c) of the line AB. We do this by substituting a point A into the general form for a straight line:

y=mx+c(1)=(0,5)(5)+cc=1,5

STEP: Write the equation of the line in the form y=mx+c
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line AB is as follows:

y=0,5x1,5

Submit your answer as:

Calculating the inclination of a line from two points

Find the inclination of the line (θ). Round your answer to 2 decimal places.

Answer: θ = °
numeric
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

To calculate the inclination of a straight line, we first need to calculate the gradient:

m=yByAxBxA=(3,5)(1,5)(4)(1)m=1

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

Now that we know the gradient, we use the following equation to determine the value of θ:

m=tan(θ)θ=tan1(m)=tan1(1)=45°

Very important: always check that your calculator is in DEG mode when using the trigonometric function buttons.


STEP: <no title>
[−0 points ⇒ 0 / 2 points left]

If the gradient of the line is negative, we add 180° to the above angle to get an obtuse angle. In this case, the gradient is positive.

θ=45°

Submit your answer as:

Calculating the inclination of a line from two points

Find the inclination of the line (θ). Round your answer to 2 decimal places.

Answer: θ = °
numeric
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

To calculate the inclination of a straight line, we first need to calculate the gradient:

m=yByAxBxA=(2,5)(0,5)(1)(1)m=1,5

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now that we know the gradient, we use the following equation to determine the value of θ:

m=tan(θ)θ=tan1(m)=tan1(1,5)=56,31°

Very important: always check that your calculator is in DEG mode when using the trigonometric function buttons.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

If the gradient of the line is negative, we add 180° to the above angle to get an obtuse angle. In this case, the gradient is negative. (You can see in the diagram that the angle of inclination is greater than 90 degrees.)

θ=56,31°+180°θ=123,69°

Submit your answer as:

Calculating the inclination of a line from two points

Find the inclination of the line (θ). Round your answer to 2 decimal places.

Answer: θ = °
numeric
STEP: <no title>
[−1 point ⇒ 1 / 2 points left]

To calculate the inclination of a straight line, we first need to calculate the gradient:

m=yByAxBxA=(3,5)(0,5)(2)(1)m=1

STEP: <no title>
[−1 point ⇒ 0 / 2 points left]

Now that we know the gradient, we use the following equation to determine the value of θ:

m=tan(θ)θ=tan1(m)=tan1(1)=45°

Very important: always check that your calculator is in DEG mode when using the trigonometric function buttons.


STEP: <no title>
[−0 points ⇒ 0 / 2 points left]

If the gradient of the line is negative, we add 180° to the above angle to get an obtuse angle. In this case, the gradient is positive.

θ=45°

Submit your answer as:

Parallel and perpendicular lines

Are the following lines parallel, perpendicular, or neither?

6=5x2y4x+10y=5

Answer choices:

  • parallel
  • perpendicular
  • neither
INSTRUCTION: Type your answer from the three choices given; mispelled answers will be marked wrong.
Answer: The lines are .
string
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To answer this question, you must know the gradients of each equation. To find the gradients, start by arranging the equations in standard form for linear equations:

y=mx+c

STEP: Arrange the first equation in standard form
[−1 point ⇒ 2 / 3 points left]

We have two equations and need to determine if they represent lines which are parallel, perpendicular or neither.

We can only answer this question by comparing the gradients of the two equations. So we need to arrange the equations in standard form. For the first equation this will be:

6=5x2y2y=5x+612(2y)=(5x+6)12y=5x2+3

The gradient of the first equation is m1=52.


STEP: Arrange the second equation in standard form
[−1 point ⇒ 1 / 3 points left]

Now change the second equation into standard form.

4x+10y=510y=4x5110(10y)=(4x5)110y=2x512

The gradient of the second equation is m2=25.


STEP: Compare the gradients of the equations
[−1 point ⇒ 0 / 3 points left]

Now we can compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those. The key facts to remember are that:

  • if the gradients are equal the lines are parallel
  • if the gradients are negative reciprocals the lines are perpendicular

For the equations in this question, the gradients are not equal, so the lines are not parallel. Also, the gradients are not negative reciprocals of each other, so the lines are not perpendicular.

The two equations are graphed on the Cartesian plane below. We can see that the lines are neither parallel nor perpendicular.

The correct answer is neither.


Submit your answer as:

Parallel and perpendicular lines

Are the lines shown in these equtions parallel, perpendicular, or neither?

2y3=2x3y=4xy+3

Answer choices:

  • parallel
  • perpendicular
  • neither
INSTRUCTION: Type your answer from the three choices given; mispelled answers will be marked wrong.
Answer: The lines are .
string
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To answer this question, you must know the gradients of each equation. To find the gradients, start by arranging the equations in standard form for linear equations:

y=mx+c

STEP: Arrange the first equation in standard form
[−1 point ⇒ 2 / 3 points left]

We have two equations and need to determine if they represent lines which are parallel, perpendicular or neither.

We can only answer this question by comparing the gradients of the two equations. So we need to arrange the equations in standard form. For the first equation this will be:

2y3=2x2y=2x+312(2y)=(2x+3)12y=x+32

The gradient of the first equation is m1=1.


STEP: Arrange the second equation in standard form
[−1 point ⇒ 1 / 3 points left]

Now change the second equation into standard form.

3y=4xy+34y=4x+314(4y)=(4x+3)14y=x+34

The gradient of the second equation is m2=1.


STEP: Compare the gradients of the equations
[−1 point ⇒ 0 / 3 points left]

Now we can compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those. The key facts to remember are that:

  • if the gradients are equal the lines are parallel
  • if the gradients are negative reciprocals the lines are perpendicular

The two gradients are negative reciprocals of each other. So the lines are perpendicular. (Notice also that the gradients have a product of 1: (1)(1) is equal to 1, which is always true for perpendicular gradients.)

The two equations are graphed on the Cartesian plane below. We can see that the two lines sit at right angles to each other: they are perpendicular.

The correct answer is perpendicular.


Submit your answer as:

Parallel and perpendicular lines

Are the lines shown in these equtions parallel, perpendicular, or neither?

20=3x5y10=3x5y

Answer choices:

  • parallel
  • perpendicular
  • neither
INSTRUCTION: Type your answer from the three choices given; mispelled answers will be marked wrong.
Answer: The lines are .
string
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

To answer this question, you must know the gradients of each equation. To find the gradients, start by arranging the equations in standard form for linear equations:

y=mx+c

STEP: Arrange the first equation in standard form
[−1 point ⇒ 2 / 3 points left]

We have two equations and need to determine if they represent lines which are parallel, perpendicular or neither.

We can only answer this question by comparing the gradients of the two equations. So we need to arrange the equations in standard form. For the first equation this will be:

20=3x5y5y=3x+2015(5y)=(3x+20)15y=3x5+4

The gradient of the first equation is m1=35.


STEP: Arrange the second equation in standard form
[−1 point ⇒ 1 / 3 points left]

Now change the second equation into standard form.

10=3x5y5y=3x1015(5y)=(3x10)15y=3x52

The gradient of the second equation is m2=35.


STEP: Compare the gradients of the equations
[−1 point ⇒ 0 / 3 points left]

Now we can compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those. The key facts to remember are that:

  • if the gradients are equal the lines are parallel
  • if the gradients are negative reciprocals the lines are perpendicular

The equations have equal gradients, so they must be parallel lines.

The two equations are graphed on the Cartesian plane below. We can see that the two lines run parallel to one another.

The correct answer is parallel.


Submit your answer as:

Calculating equations of perpendicular lines

Consider the following diagram:

Line AB is perpendicular to the line y=2x1,25 (blue line).

Determine the equation of AB.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of a straight line equation is y=mx+c.

To calculate the equation of the straight line, we first need the gradient (m) of AB, which we can get from the perpendicular line:

For perpendicular lines, we know that m1×m2=1:

mAB×2=1mAB=12=0,5

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we substitute the coordinates of A into the equation to calculate the value of the y-intercept (c):

y=0,5x+c3=(0,5)(3)+cc=1,5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is:

y=0,5x1,5

Submit your answer as:

Calculating equations of perpendicular lines

Consider the following diagram:

Line AB is perpendicular to the line y=x1,0 (blue line).

Find the equation of AB.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of a straight line equation is y=mx+c.

To calculate the equation of the straight line, we first need the gradient (m) of AB, which we can get from the perpendicular line:

For perpendicular lines, we know that m1×m2=1:

mAB×1=1mAB=11=1

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we substitute the coordinates of A into the equation to calculate the value of the y-intercept (c):

y=1x+c1,5=(1)(2)+cc=0,5

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is:

y=x0,5

Submit your answer as:

Calculating equations of perpendicular lines

Consider the following diagram:

Line AB is perpendicular to the line y=2x+0,75 (blue line).

Find the equation of AB.

Answer: y =
expression
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

The general form of a straight line equation is y=mx+c.

To calculate the equation of the straight line, we first need the gradient (m) of AB, which we can get from the perpendicular line:

For perpendicular lines, we know that m1×m2=1:

mAB×2=1mAB=12=0,5

STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Next we substitute the coordinates of A into the equation to calculate the value of the y-intercept (c):

y=0,5x+c0=(0,5)(2)+cc=1

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of AB is:

y=0,5x+1

Submit your answer as:

Using inclination to find the equation of a line

The line shown below has an angle of inclination θ = 153,43°.

Determine the equation of the line. Round your answer to one decimal place.

Answer:y =
expression
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

We are given the angle of inclination of the line, which we can use to calculate the gradient:

m=tanθ=tan(153,43°)=0,5

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

We know the general form of the equation of a straight line is y=mx+c.

Substitute the value for m into the equation:

y=0,5x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

Next we calculate the value of c by substituting the coordinates of point A(2;1) into the equation and simplifying:

y=0,5x+c1=(0,5)(2)+cc=2

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Therefore, the equation of the line AB is

y=0,5x2

Submit your answer as:

Using inclination to find the equation of a line

The line shown below has an angle of inclination θ = 116,57°.

Find the equation of the line. Round your answer to one decimal place.

Answer:y =
expression
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

We are given the angle of inclination of the line, which we can use to calculate the gradient:

m=tanθ=tan(116,57°)=2

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

We know the general form of the equation of a straight line is y=mx+c.

Substitute the value for m into the equation:

y=2x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

Next we calculate the value of c by substituting the coordinates of point A(1;2.5) into the equation and simplifying:

y=2x+c2,5=(2)(1)+cc=0,5

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Therefore, the equation of the line AB is

y=2x+0,5

Submit your answer as:

Using inclination to find the equation of a line

The line shown below has an angle of inclination θ = 123,69°.

Determine the equation of the line. Round your answer to one decimal place.

Answer:y =
expression
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

We are given the angle of inclination of the line, which we can use to calculate the gradient:

m=tanθ=tan(123,69°)=1,5

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

We know the general form of the equation of a straight line is y=mx+c.

Substitute the value for m into the equation:

y=1,5x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

Next we calculate the value of c by substituting the coordinates of point A(2;2) into the equation and simplifying:

y=1,5x+c2=(1,5)(2)+cc=1

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Therefore, the equation of the line AB is

y=1,5x1

Submit your answer as:

Finding the equation of a line

Determine the equation of the line which passes through the point (1;0) and is parallel to the line y=2x+2.

Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.

Answer:
equation
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. From the given equation you can see that the gradient is m=2.

For lines that are parallel, we know that the gradients are equal so m=2.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Substitute the coordinates (1;0) and the gradient into the equation to find the value of c.

y=mx+c0=(2)(1)+c0=2+c2=c

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is y=2x2.

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (1;0) is shown with a black dot labelled with an A. You can see that the red line passes through the point and that the two lines are parallel.


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Finding the equation of a line

Determine the equation of the line which passes through the point (3;3) and is parallel to the line y=4x3+3.

Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.

Answer:
equation
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. From the given equation you can see that the gradient is m=43.

For lines that are parallel, we know that the gradients are equal so m=43.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Substitute the coordinates (3;3) and the gradient into the equation to find the value of c.

y=mx+c3=(43)(3)+c3=4+c7=c

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is y=4x3+7.

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (3;3) is shown with a black dot labelled with an A. You can see that the red line passes through the point and that the two lines are parallel.


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Finding the equation of a line

Determine the equation of the line which passes through the point (2;6) and is perpendicular to the line y=x24.

Type the entire equation in the answer box, which will either be of the form y=mx+c or x=c.

Answer:
equation
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To find the equation of any straight line, you need a point on the line (which is given in the question) and the gradient of the line. Start by finding the gradient of the line.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Start by finding the gradient of the line. From the given equation you can see that the gradient is m=12.

For lines that are perpendicular, the gradient must be the negative reciprocal, therefore m=2.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Substitute the coordinates (2;6) and the gradient into the equation to find the value of c.

y=mx+c6=(2)(2)+c6=4+c10=c

STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

Therefore, the equation of the line is y=2x+10.

The two equations are graphed on the Cartesian plane shown below: the blue line is the equation given in the question, and the red line is the answer. The point (2;6) is shown with a black dot labelled with an A. You can see that the red line passes through the point and that the lines are perpendicular to each other.


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Finding a coordinate of a point on a perpendicular line

You are given the following diagram, which shows two perpendicular lines on the Cartesian plane. Line AB has endpoints at A(1;2) and B(3;y). The other line (the dashed line) has the equation y=x.

Calculate the missing value of y.

Answer:

y=

numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The gradient of the line AB is equal to:

1gradientPL

where PL stands for Perpendicular Line.


STEP: Write down the equation of a straight line
[−1 point ⇒ 3 / 4 points left]

We need to find the equation of line AB. So we will need the equation of a straight line:

y=mx+c

STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 4 points left]

The line AB has a gradient that is equal to:

1gradientPL

where PL stands for Perpendicular Line. The gradient of the perpendicular line is 1, so:

mAB=1gradientPL=1(1)=1

Now we know that the equation of line AB is

y=x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known point A into the equation, to find c:

y=x+c(2)=(1)+cc=1

And now we have the complete equation for line AB:

y=x+1

STEP: Substitute the known value for point B into the equation and solve
[−1 point ⇒ 0 / 4 points left]

We can use the equation for line AB to find the missing coordinate. To do this, substitute the known value for point B into the equation for line AB:

y=x+1y=(3)+1

Solving, we get:

y=2

The correct answer is y=2.


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Finding a coordinate of a point on a perpendicular line

You are given the following diagram, which shows two perpendicular lines on the Cartesian plane. Line AB has endpoints at A(3;72) and B(x;72). The other line (the dashed line) has the equation y=x.

Calculate the missing value of x.

Answer:

x=

numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The gradient of the line AB is equal to:

1gradientPL

where PL stands for Perpendicular Line.


STEP: Write down the equation of a straight line
[−1 point ⇒ 3 / 4 points left]

We need to find the equation of line AB. So we will need the equation of a straight line:

y=mx+c

STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 4 points left]

The line AB has a gradient that is equal to:

1gradientPL

where PL stands for Perpendicular Line. The gradient of the perpendicular line is 1, so:

mAB=1gradientPL=1(1)=1

Now we know that the equation of line AB is

y=x+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known point A into the equation, to find c:

y=x+c(72)=(3)+cc=12

And now we have the complete equation for line AB:

y=x+12

STEP: Substitute the known value for point B into the equation and solve
[−1 point ⇒ 0 / 4 points left]

We can use the equation for line AB to find the missing coordinate. To do this, substitute the known value for point B into the equation for line AB:

y=x+12(72)=x+12

Solving, we get:

x=4

The correct answer is x=4.


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Finding a coordinate of a point on a perpendicular line

You are given the following diagram, which shows two perpendicular lines on the Cartesian plane. Line AB has endpoints at A(1;1) and B(x;52). The other line (the dashed line) has the equation y=2x74.

Calculate the missing value of x.

Answer:

x=

numeric
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

The gradient of the line AB is equal to:

1gradientPL

where PL stands for Perpendicular Line.


STEP: Write down the equation of a straight line
[−1 point ⇒ 3 / 4 points left]

We need to find the equation of line AB. So we will need the equation of a straight line:

y=mx+c

STEP: Determine the gradient of line AB using the gradient of the perpendicular line
[−1 point ⇒ 2 / 4 points left]

The line AB has a gradient that is equal to:

1gradientPL

where PL stands for Perpendicular Line. The gradient of the perpendicular line is 2, so:

mAB=1gradientPL=1(2)=12

Now we know that the equation of line AB is

y=x2+c

STEP: Calculate the y-intercept of line AB
[−1 point ⇒ 1 / 4 points left]

Now we substitute the known point A into the equation, to find c:

y=x2+c(1)=(1)2+cc=32

And now we have the complete equation for line AB:

y=x232

STEP: Substitute the known value for point B into the equation and solve
[−1 point ⇒ 0 / 4 points left]

We can use the equation for line AB to find the missing coordinate. To do this, substitute the known value for point B into the equation for line AB:

y=x232(52)=x232

Solving, we get:

x=2

The correct answer is x=2.


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Parallel and perpendicular lines

Determine whether the equations represent lines which are parallel, perpendicular, or neither.

4y4=4x13=4x4y
Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, we need to know the gradient of each equation. This means that we need to start by arranging the equations in standard form for linear equations: y=mx+c.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

We answer this question by comparing the gradients of the two equations, for which the equations have to be in standard form. For the first equation this is:

4y4=4x14y=4x+314(4y)=14(4x+3)y=x+34

The gradient of the first equation is m=1.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now we change the second equation into standard form.

3=4x4y4y=4x314(4y)=14(4x3)y=x34

The gradient of the second equation is m=1.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

We now compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those.

  • Are the gradients equal to each other (parallel gradients)?
  • Do they have a product of 1 (perpendicular gradients)?
  • Are neither of these true?

The two gradients have a product of 1: m1×m2=(1)(1)=1.
Therefore, the lines are perpendicular. The correct response is: perpendicular.

The two equations are graphed on the Cartesian plane shown below. We can see that the two lines sit at right angles to each other: they are perpendicular.


Submit your answer as:

Parallel and perpendicular lines

Are the following lines parallel, perpendicular, or neither?

8x+12y+9=02y4=10x7
Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, we need to know the gradient of each equation. This means that we need to start by arranging the equations in standard form for linear equations: y=mx+c.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

We answer this question by comparing the gradients of the two equations, for which the equations have to be in standard form. For the first equation this is:

8x+12y+9=012y=8x9112(12y)=112(8x9)y=2x334

The gradient of the first equation is m=23.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now we change the second equation into standard form.

2y4=10x72y=10x312(2y)=12(10x3)y=5x32

The gradient of the second equation is m=5.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

We now compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those.

  • Are the gradients equal to each other (parallel gradients)?
  • Do they have a product of 1 (perpendicular gradients)?
  • Are neither of these true?

For the equations in this question, the gradients are not equal, so the lines cannot be parallel. Also, the gradients are not negative reciprocals of each other, so the lines cannot be perpendicular. The correct response is: neither.

The two equations are graphed on the Cartesian plane shown below. We can see that the lines are neither parallel nor perpendicular.


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Parallel and perpendicular lines

Investigate whether these lines are parallel, perpendicular, or neither.

8x=4y13x=3y+4
Answer: The lines are
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]
To answer this question, we need to know the gradient of each equation. This means that we need to start by arranging the equations in standard form for linear equations: y=mx+c.
STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

We answer this question by comparing the gradients of the two equations, for which the equations have to be in standard form. For the first equation this is:

8x=4y14y=8x114(4y)=14(8x1)y=2x14

The gradient of the first equation is m=2.


STEP: <no title>
[−1 point ⇒ 1 / 3 points left]

Now we change the second equation into standard form.

3x=3y+43y=3x+413(3y)=13(3x+4)y=x+43

The gradient of the second equation is m=1.


STEP: <no title>
[−1 point ⇒ 0 / 3 points left]

We now compare the two gradients to decide if the lines are parallel, perpendicular, or neither of those.

  • Are the gradients equal to each other (parallel gradients)?
  • Do they have a product of 1 (perpendicular gradients)?
  • Are neither of these true?

For the equations in this question, the gradients are not equal, so the lines cannot be parallel. Also, the gradients are not negative reciprocals of each other, so the lines cannot be perpendicular. The correct response is: neither.

The two equations are graphed on the Cartesian plane shown below. We can see that the lines are neither parallel nor perpendicular.


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Calculating inclination from two points

Consider the diagram below, which shows a line through the points: A(5;203) and B(3;4).

Find the inclination of the line, which is the angle θ. Round your answer to 2 decimal places if appropriate.

Answer: θ = °
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need the inclination of the line, which is the angle θ. The inclination is related to the gradient, so start by finding the gradient of the line.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

We need to find the inclination of the line. The inclination is the angle between the line and the positive x-axis. This is shown by the angle θ in the diagram.

The inclination of a line is directly related to the line's gradient: both represent the steepness of the line. (For an explanation about the connection between the gradient and the inclination, you can read here.) So to find the inclination, we need the gradient of the line. Be aware that there is a double negative in the top of the fraction, since point A has a negative y coordinate.

m=y2y1x2x1=(4)(203)(3)(5)=4+20335=43

STEP: Calculate the inclination using the inclination formula
[−1 point ⇒ 1 / 3 points left]

Now we can calculate the inclination of the line. We do this using the following equation:

m=tanθθ=tan1(m)=tan1(43)=53,13010...°53,13°

STEP: Adjust the answer if necessary
[−1 point ⇒ 0 / 3 points left]

If the gradient is negative we must add 180° to the above value to get an obtuse angle. In this case, the gradient, and the angle we got above, is negative. (In fact, you can see in the diagram that the angle of inclination is greater than 90 degrees. For an explanation of why we must add 180° to the answer when the gradient is negative, see this information about inclination .)

θnegative53,13°+180°θ126,87°

The inclination of the line is 126,87°.


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Calculating inclination from two points

Consider the diagram below, which shows a line through the points: A(5;214) and B(8;92).

Find the inclination of the line, which is the angle θ. Round your answer to 2 decimal places if appropriate.

Answer: θ = °
numeric
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need the inclination of the line, which is the angle θ. The inclination is related to the gradient, so start by finding the gradient of the line.


STEP: Find the gradient of the line
[−1 point ⇒ 1 / 2 points left]

We need to find the inclination of the line. The inclination is the angle between the line and the positive x-axis. This is shown by the angle θ in the diagram.

The inclination of a line is directly related to the line's gradient: both represent the steepness of the line. (For an explanation about the connection between the gradient and the inclination, you can read here.) So to find the inclination, we need the gradient of the line. Watch out: there are double negatives in the numerator and the denominator in this gradient calculation.

m=y2y1x2x1=(92)(214)(8)(5)=92+2148+5=34

STEP: Calculate the inclination using the inclination formula
[−1 point ⇒ 0 / 2 points left]

Now we can calculate the inclination of the line. We do this using the following equation:

m=tanθθ=tan1(m)=tan1(34)=36,86989...°36,87°

STEP: Adjust the answer if necessary
[−0 points ⇒ 0 / 2 points left]

If the gradient is negative we must add 180° to the above value to get an obtuse angle. In this case, the gradient, and the angle we got above, is positive.

θ36,87°

The inclination of the line is 36,87°.


Submit your answer as:

Calculating inclination from two points

Consider the diagram below, which shows a line through the points: A(2;92) and B(7;92).

Find the inclination of the line, which is the angle θ. Round your answer to 2 decimal places if appropriate.

Answer: θ = °
numeric
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

You need the inclination of the line, which is the angle θ. The inclination is related to the gradient, so start by finding the gradient of the line.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

We need to find the inclination of the line. The inclination is the angle between the line and the positive x-axis. This is shown by the angle θ in the diagram.

The inclination of a line is directly related to the line's gradient: both represent the steepness of the line. (For an explanation about the connection between the gradient and the inclination, you can read here.) So to find the inclination, we need the gradient of the line. Be aware that there is a double negative in the top of the fraction, since point A has a negative y coordinate.

m=y2y1x2x1=(92)(92)(7)(2)=92+9272=1

STEP: Calculate the inclination using the inclination formula
[−1 point ⇒ 1 / 3 points left]

Now we can calculate the inclination of the line. We do this using the following equation:

m=tanθθ=tan1(m)=tan1(1)=45°

STEP: Adjust the answer if necessary
[−1 point ⇒ 0 / 3 points left]

If the gradient is negative we must add 180° to the above value to get an obtuse angle. In this case, the gradient, and the angle we got above, is negative. (In fact, you can see in the diagram that the angle of inclination is greater than 90 degrees. For an explanation of why we must add 180° to the answer when the gradient is negative, see this information about inclination .)

θnegative=45°+180°θ=135°

The inclination of the line is 135°.


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Working with perpendicular lines

In the diagram below, the two lines are perpendicular. The equation of the dashed red line is

y=3x21

As labelled below, the solid blue line passes through the points A(5;13) and B(3;y).

Determine yB, the y-coordinate of point B.

Answer: yB =
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by finding the gradient of line AB. You can do this because the lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 4 / 5 points left]

We need to find the equation of the blue line, which is called line AB. That means we need the gradient and the y-intercept of line AB.

Since the two lines in the figure are perpendicular to each other, we can find the gradient of line AB from the gradient of the dashed red line. Perpendicular gradients are negative reciprocals of each other.

The gradient of the dashed red line is 32 (we can read it straight from the equation), so:

mAB=negative reciprocal of mred line=reciprocal of 32=(23)=23

STEP: Start building the equation
[−1 point ⇒ 3 / 5 points left]

Since the equation for a staight line is:

y=mx+c

The equation for line AB must be:

y=23x+c

STEP: Find the y-intercept of the equation
[−1 point ⇒ 2 / 5 points left]

Now we substitute the coordinates of point A into the equation, to find c:

y=23x+c13=23(5)+c3=c

Now we know the equation for line AB:

y=2x33

STEP: Use the x-coordinate of point B to find yB
[−2 points ⇒ 0 / 5 points left]

Now we use a key concept for graphs and their equations: every point on the line agrees with the equation. If we substitute in the x-coordinate of any point on the line into the equation, we will always get the y-coordinate for that point.The equation forces this to be true! So now substitute the x-coordinate for point B into the equation for line AB and solve for yB.

yB=2xB33yB=2(3)33yB=5

The correct answer is yB=5.


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Working with perpendicular lines

In the diagram below, the two lines are perpendicular. The equation of the dashed red line is

y=4x3+72

As labelled below, the solid blue line passes through the points A(6;3) and B(x;154).

Determine xB, the x-coordinate of point B.

Answer: xB =
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by finding the gradient of line AB. You can do this because the lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 4 / 5 points left]

We need to find the equation of the blue line, which is called line AB. That means we need the gradient and the y-intercept of line AB.

Since the two lines in the figure are perpendicular to each other, we can find the gradient of line AB from the gradient of the dashed red line. Perpendicular gradients are negative reciprocals of each other.

The gradient of the dashed red line is 43 (we can read it straight from the equation), so:

mAB=negative reciprocal of mred line=reciprocal of 43=(34)=34

STEP: Start building the equation
[−1 point ⇒ 3 / 5 points left]

Since the equation for a staight line is:

y=mx+c

The equation for line AB must be:

y=34x+c

STEP: Find the y-intercept of the equation
[−1 point ⇒ 2 / 5 points left]

Now we substitute the coordinates of point A into the equation, to find c:

y=34x+c3=34(6)+c32=c

Now we know the equation for line AB:

y=3x432

STEP: Use the y-coordinate of point B to find xB
[−2 points ⇒ 0 / 5 points left]

Now we use a key concept for graphs and their equations: every point on the line agrees with the equation. If we substitute in the y-coordinate of any point on the line into the equation, we will always get the x-coordinate for that point.The equation forces this to be true! So now substitute the y-coordinate for point B into the equation for line AB and solve for xB.

yB=3xB432(154)=3xB432xB=3

The correct answer is xB=3.


Submit your answer as:

Working with perpendicular lines

In the diagram below, the two lines are perpendicular. The equation of the dashed red line is

y=2x372

As labelled below, the solid blue line passes through the points A(5;7) and B(x;5).

Determine xB, the x-coordinate of point B.

Answer: xB =
numeric
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by finding the gradient of line AB. You can do this because the lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 4 / 5 points left]

We need to find the equation of the blue line, which is called line AB. That means we need the gradient and the y-intercept of line AB.

Since the two lines in the figure are perpendicular to each other, we can find the gradient of line AB from the gradient of the dashed red line. Perpendicular gradients are negative reciprocals of each other.

The gradient of the dashed red line is 23 (we can read it straight from the equation), so:

mAB=negative reciprocal of mred line=reciprocal of 23=(32)=32

STEP: Start building the equation
[−1 point ⇒ 3 / 5 points left]

Since the equation for a staight line is:

y=mx+c

The equation for line AB must be:

y=32x+c

STEP: Find the y-intercept of the equation
[−1 point ⇒ 2 / 5 points left]

Now we substitute the coordinates of point A into the equation, to find c:

y=32x+c7=32(5)+c12=c

Now we know the equation for line AB:

y=3x2+12

STEP: Use the y-coordinate of point B to find xB
[−2 points ⇒ 0 / 5 points left]

Now we use a key concept for graphs and their equations: every point on the line agrees with the equation. If we substitute in the y-coordinate of any point on the line into the equation, we will always get the x-coordinate for that point.The equation forces this to be true! So now substitute the y-coordinate for point B into the equation for line AB and solve for xB.

yB=3xB2+12(5)=3xB2+12xB=3

The correct answer is xB=3.


Submit your answer as:

Facts about perpendicular lines

  1. From the following table, which statement or statements are true about perpendicular lines and the gradients of perpendicular lines? Identify all of the true statements.

    A m1×m2=1
    B Perpendicular gradients are negative reciprocals of each other.
    C Perpendicular gradients have a sum of -1.
    D m1×m2=1
    INSTRUCTION: There may be more than one true statement above. You need to identify all true statements.
    Answer: Statement(s) is (are) true
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    There are two true statements in the table above.
    STEP: Use facts about perpendicular lines
    [−1 point ⇒ 0 / 1 points left]

    The following statements are true for the gradients of perpendicular lines:

    • The gradients must be negative reciprocals of each other. For example, the negative reciprocal of 3 is 13.
    • The gradients must have a product of -1 : m1×m2=1. For example, 3 and 13 are perpendicular gradients because
      (3)×(13)=1
      .

    There are two statements in the table above which agree with these facts.

    The correct choice is: B and D.


    Submit your answer as:
  2. If a certain line has a gradient of m=54, what is the gradient of a line perpendicular to that line? (Remember that division by zero is 'undefined.')

    Answer: m=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the facts from Question 1 to work out the answer!


    STEP: Find the negative reciprocal of the given gradient
    [−1 point ⇒ 0 / 1 points left]

    If the gradients are perpendicular, then the gradient we want must be the negative reciprocal of the gradient given. 'Negative' means change the sign (multiply by -1) and 'reciprocal' means turn the number upside down. The gradient given is m=54: the negative of that is 54. Then the reciprocal of that is 45.

    Note: for perpendicular gradients, m1×m2=1. The gradient given is m=54, so we want to solve (54)×m2=1. The value which will solve that equation is 45.

    Finally, we can look at a graph of lines which have the gradients above. You can see that they are perpendicular to each other! You can also see that one of the lines has a positive gradient while the other one has a negative gradient - they are negatives.

    The correct answer is 45.


    Submit your answer as:

Facts about perpendicular lines

  1. From the following table, which statement or statements are true about perpendicular lines and the gradients of perpendicular lines? Identify all of the true statements.

    A Perpendicular gradients are equal.
    B m1×m2=1
    C Perpendicular gradients are negative reciprocals of each other.
    D Perpendicular gradients have a product of -1.
    INSTRUCTION: There may be more than one true statement above. You need to identify all true statements.
    Answer: Statement(s) is (are) true
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    There are two true statements in the table above.
    STEP: Use facts about perpendicular lines
    [−1 point ⇒ 0 / 1 points left]

    The following statements are true for the gradients of perpendicular lines:

    • The gradients must be negative reciprocals of each other. For example, the negative reciprocal of 12 is 2.
    • The gradients must have a product of -1 : m1×m2=1. For example, 12 and 2 are perpendicular gradients because
      (12)×(2)=1
      .

    There are two statements in the table above which agree with these facts.

    The correct choice is: C and D.


    Submit your answer as:
  2. If a certain line has a gradient of m=53, what is the gradient of a line perpendicular to that line? (Remember that division by zero is 'undefined.')

    Answer: m=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the facts from Question 1 to work out the answer!


    STEP: Find the negative reciprocal of the given gradient
    [−1 point ⇒ 0 / 1 points left]

    If the gradients are perpendicular, then the gradient we want must be the negative reciprocal of the gradient given. 'Negative' means change the sign (multiply by -1) and 'reciprocal' means turn the number upside down. The gradient given is m=53: the negative of that is 53. Then the reciprocal of that is 35.

    Note: for perpendicular gradients, m1×m2=1. The gradient given is m=53, so we want to solve (53)×m2=1. The value which will solve that equation is 35.

    Finally, we can look at a graph of lines which have the gradients above. You can see that they are perpendicular to each other! You can also see that one of the lines has a positive gradient while the other one has a negative gradient - they are negatives.

    The correct answer is 35.


    Submit your answer as:

Facts about perpendicular lines

  1. From the following table, which statement or statements are true about perpendicular lines and the gradients of perpendicular lines? Identify all of the true statements.

    A Perpendicular gradients have a sum of -1.
    B Perpendicular gradients have a product of -1.
    C Perpendicular gradients are negative reciprocals of each other.
    D m1×m2=1
    INSTRUCTION: There may be more than one true statement above. You need to identify all true statements.
    Answer: Statement(s) is (are) true
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    There are two true statements in the table above.
    STEP: Use facts about perpendicular lines
    [−1 point ⇒ 0 / 1 points left]

    The following statements are true for the gradients of perpendicular lines:

    • The gradients must be negative reciprocals of each other. For example, the negative reciprocal of 1 is 1.
    • The gradients must have a product of -1 : m1×m2=1. For example, 1 and 1 are perpendicular gradients because
      (1)×(1)=1
      .

    There are two statements in the table above which agree with these facts.

    The correct choice is: B and C.


    Submit your answer as:
  2. If a certain line has a gradient of m=23, what is the gradient of a line perpendicular to that line? (Remember that division by zero is 'undefined.')

    Answer: m=
    numeric
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use the facts from Question 1 to work out the answer!


    STEP: Find the negative reciprocal of the given gradient
    [−1 point ⇒ 0 / 1 points left]

    If the gradients are perpendicular, then the gradient we want must be the negative reciprocal of the gradient given. 'Negative' means change the sign (multiply by -1) and 'reciprocal' means turn the number upside down. The gradient given is m=23: the negative of that is 23. Then the reciprocal of that is 32.

    Note: for perpendicular gradients, m1×m2=1. The gradient given is m=23, so we want to solve (23)×m2=1. The value which will solve that equation is 32.

    Finally, we can look at a graph of lines which have the gradients above. You can see that they are perpendicular to each other! You can also see that one of the lines has a positive gradient while the other one has a negative gradient - they are negatives.

    The correct answer is 32.


    Submit your answer as:

Finding the equation of a perpendicular line

The figure below shows two perpendicular lines. The equation of the dashed red line is:

y=2x52

The solid blue line passes through points A and B. The coordinates of point B are (4;92) and the coordinates of A are unknown.

Determine the equation of line AB.

INSTRUCTION: Type only the right side of the equation into the answer box (following the "y = ...").
Answer: Equation of line AB: y =
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the gradient of line AB. You can do this using the fact that the two lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

The equation of a straight line requires the gradient and the y-intercept for the line. We can find the gradient in this case because line AB is perpendicular to the dashed red line: if we know a gradient, we can always find the gradient which is perpendicular to it.

The equation of the dashed red line is

y=2x52

which means that the gradient of the dashed line is 2.

Perpendicular lines have gradients which are negative reciprocals of each other. So the gradient of line AB is the negative reciprocal of 2.

mAB=negative reciprocal of mdashed line=reciprocal of 2=(12)=12

This means that the equation of line AB is:

y=x2+c

STEP: Find the value of c
[−1 point ⇒ 1 / 3 points left]

Now we can find the value of c for the line (which is the y-intercept). We do this by substituting any point on the line into the equation from above. We only know the coordinates of point B, which are (4;92). Substitute these coordinates into the equation and solve for c.

y=x2+c(92)=(4)2+ccoordinates of point BSubstitute in the52=c

STEP: Write the equation
[−1 point ⇒ 0 / 3 points left]

We now have both numbers we need for the equation of line AB: we just need to put them together.

The correct answer is:

y=x2+52

Submit your answer as:

Finding the equation of a perpendicular line

The figure below shows two perpendicular lines. The equation of the dashed red line is:

y=x272

The solid blue line passes through points A and B. The coordinates of point A are (1;32) and the coordinates of B are unknown.

Determine the equation of line AB.

INSTRUCTION: Type only the right side of the equation into the answer box (following the "y = ...").
Answer: Equation of line AB: y =
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the gradient of line AB. You can do this using the fact that the two lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

The equation of a straight line requires the gradient and the y-intercept for the line. We can find the gradient in this case because line AB is perpendicular to the dashed red line: if we know a gradient, we can always find the gradient which is perpendicular to it.

The equation of the dashed red line is

y=x272

which means that the gradient of the dashed line is 12.

Perpendicular lines have gradients which are negative reciprocals of each other. So the gradient of line AB is the negative reciprocal of 12.

mAB=negative reciprocal of mdashed line=reciprocal of 12=(2)=2

This means that the equation of line AB is:

y=2x+c

STEP: Find the value of c
[−1 point ⇒ 1 / 3 points left]

Now we can find the value of c for the line (which is the y-intercept). We do this by substituting any point on the line into the equation from above. We only know the coordinates of point A, which are (1;32). Substitute these coordinates into the equation and solve for c.

y=2x+c(32)=2(1)+ccoordinates of point ASubstitute in the12=c

STEP: Write the equation
[−1 point ⇒ 0 / 3 points left]

We now have both numbers we need for the equation of line AB: we just need to put them together.

The correct answer is:

y=2x12

Submit your answer as:

Finding the equation of a perpendicular line

The figure below shows two perpendicular lines. The equation of the dashed red line is:

y=x25

The solid blue line passes through points A and B. The coordinates of point B are (3;7) and the coordinates of A are unknown.

Calculate the equation of line AB.

INSTRUCTION: Type only the right side of the equation into the answer box (following the "y = ...").
Answer: Equation of line AB: y =
expression
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by finding the gradient of line AB. You can do this using the fact that the two lines are perpendicular.


STEP: Find the gradient of the line
[−1 point ⇒ 2 / 3 points left]

The equation of a straight line requires the gradient and the y-intercept for the line. We can find the gradient in this case because line AB is perpendicular to the dashed red line: if we know a gradient, we can always find the gradient which is perpendicular to it.

The equation of the dashed red line is

y=x25

which means that the gradient of the dashed line is 12.

Perpendicular lines have gradients which are negative reciprocals of each other. So the gradient of line AB is the negative reciprocal of 12.

mAB=negative reciprocal of mdashed line=reciprocal of 12=(2)=2

This means that the equation of line AB is:

y=2x+c

STEP: Find the value of c
[−1 point ⇒ 1 / 3 points left]

Now we can find the value of c for the line (which is the y-intercept). We do this by substituting any point on the line into the equation from above. We only know the coordinates of point B, which are (3;7). Substitute these coordinates into the equation and solve for c.

y=2x+c(7)=2(3)+ccoordinates of point BSubstitute in the1=c

STEP: Write the equation
[−1 point ⇒ 0 / 3 points left]

We now have both numbers we need for the equation of line AB: we just need to put them together.

The correct answer is:

y=2x1

Submit your answer as:

Finding a point using perpendicular lines

The following diagram is given:

Line AB is perpendicular to the line y=x+1,5.

Determine the x-coordinate of point B, rounded to 2 decimal places.

Answer: B (;2,5)
numeric
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Recall the standard form of a straight line equation:

y=mx+c

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

For perpendicular lines, we know that m1×m2=1:

mAB×1=1mAB=11=1y=1x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

We substitute the x and y coordinates of point A into the equation to find c:

y=1x+c0,5=(1)(1)+cc=1,5

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Finally, we substitute the y-coordinate for point B into the equation and solve for x:

y=1x+1,52,5=(1)(x)+1,5x=1

Submit your answer as:

Finding a point using perpendicular lines

The following diagram is given:

Line AB is perpendicular to the line y=x+2.

Find the y-coordinate of point B, rounded to 2 decimal places.

Answer: B (2;)
numeric
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Recall the standard form of a straight line equation:

y=mx+c

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

For perpendicular lines, we know that m1×m2=1:

mAB×1=1mAB=11=1y=1x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

We substitute the x and y coordinates of point A into the equation to find c:

y=1x+c0,5=(1)(1)+cc=1,5

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Finally, we substitute the x-coordinate for point B into the equation and solve for y:

y=1x+1,5y=(1)(2)+1,5y=3,5

Submit your answer as:

Finding a point using perpendicular lines

The following diagram is given:

Line AB is perpendicular to the line y=0,5x.

Calculate the y-coordinate of point B, rounded to 2 decimal places.

Answer: B (1;)
numeric
STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Recall the standard form of a straight line equation:

y=mx+c

STEP: <no title>
[−1 point ⇒ 2 / 4 points left]

For perpendicular lines, we know that m1×m2=1:

mAB×0,5=1mAB=10,5=2y=2x+c

STEP: <no title>
[−1 point ⇒ 1 / 4 points left]

We substitute the x and y coordinates of point A into the equation to find c:

y=2x+c4=(2)(3)+cc=2

STEP: <no title>
[−1 point ⇒ 0 / 4 points left]

Finally, we substitute the x-coordinate for point B into the equation and solve for y:

y=2x2y=(2)(1)2y=4

Submit your answer as:

Finding the equation of a line from the inclination

The diagram below shows a line on the Cartesian plane. The line passes through the point A(5;6). The line has an angle of inclination θ=123,69°.

Find the equation of the line.

INSTRUCTIONS:
  • Give only the right side of the equation, following y=....
  • Round the values in the equation to 2 decimal places if necessary.
Answer: Complete the equation for the line: y = .
one-of
type(expression)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the inclination to find the gradient of the line. Then use that gradient to start working out the equation.


STEP: Use the inclination to find the gradient of the line
[−1 point ⇒ 3 / 4 points left]

This question is about finding the equation of a straight line. We know one point on the line and the line's inclination. The equation of the line will have the form y=mx+c. So we need to find the gradient, m, and the y-intercept, c.

The inclination of a line tells us about the steepness of the line so it is similar to the gradient. In fact, inclination and gradient are directly related to each other. We can use the following equation to calculate the gradient from the inclination.

m=tanθ=tan(123,69°)=1,50000...1,5

The gradient is approximately 1,5 (remember that the instructions state the we should round the values to 2 decimal places). Note that the gradient is negative, which agrees with the slope of the graph.


STEP: Put the gradient into the equation
[−1 point ⇒ 2 / 4 points left]

Now we recall the general form of the equation of a straight line. Since we know the gradient m of the line, we can substitute that into the equation of the line:

y=(1,5)x+c

STEP: Use the point to find the y-intercept
[−1 point ⇒ 1 / 4 points left]

Now we can calculate the value of the y-intercept of the line (the value of c). We do this by substituting the coordinates of any point on the line into the general form for the line from above. Since we know the coordinates of point A, we will use those coordinates, which are A(5;6).

y=1,5x+c(6)=1,5(5)+c1,5=c

Remember that this number should be the y-intercept of the graph. As with the gradient, we can compare this answer to the graph to check if they agree: does the graph up above have a y-intercept close to 1,5? Yes!


STEP: Write the complete equation
[−1 point ⇒ 0 / 4 points left]

We now have the gradient and the y-intercept of the line. The equation of the line is:

y=1,5x1,5

Submit your answer as:

Finding the equation of a line from the inclination

The diagram below shows a line on the Cartesian plane. The line passes through the point A(3;112). The line has an angle of inclination θ=123,69°.

Find the equation of the line.

INSTRUCTIONS:
  • Give only the right side of the equation, following y=....
  • Round the values in the equation to 2 decimal places if necessary.
Answer: Complete the equation for the line: y = .
one-of
type(expression)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the inclination to find the gradient of the line. Then use that gradient to start working out the equation.


STEP: Use the inclination to find the gradient of the line
[−1 point ⇒ 3 / 4 points left]

This question is about finding the equation of a straight line. We know one point on the line and the line's inclination. The equation of the line will have the form y=mx+c. So we need to find the gradient, m, and the y-intercept, c.

The inclination of a line tells us about the steepness of the line so it is similar to the gradient. In fact, inclination and gradient are directly related to each other. We can use the following equation to calculate the gradient from the inclination.

m=tanθ=tan(123,69°)=1,50000...1,5

The gradient is approximately 1,5 (remember that the instructions state the we should round the values to 2 decimal places). Note that the gradient is negative, which agrees with the slope of the graph.


STEP: Put the gradient into the equation
[−1 point ⇒ 2 / 4 points left]

Now we recall the general form of the equation of a straight line. Since we know the gradient m of the line, we can substitute that into the equation of the line:

y=(1,5)x+c

STEP: Use the point to find the y-intercept
[−1 point ⇒ 1 / 4 points left]

Now we can calculate the value of the y-intercept of the line (the value of c). We do this by substituting the coordinates of any point on the line into the general form for the line from above. Since we know the coordinates of point A, we will use those coordinates, which are A(3;112).

y=1,5x+c(112)=1,5(3)+c1=c

Remember that this number should be the y-intercept of the graph. As with the gradient, we can compare this answer to the graph to check if they agree: does the graph up above have a y-intercept close to 1? Yes!


STEP: Write the complete equation
[−1 point ⇒ 0 / 4 points left]

We now have the gradient and the y-intercept of the line. The equation of the line is:

y=1,5x+1

Submit your answer as:

Finding the equation of a line from the inclination

The diagram below shows a line on the Cartesian plane. The line passes through the point A(3;112). The line has an angle of inclination θ=116,565°.

Determine the equation of the line.

INSTRUCTIONS:
  • Give only the right side of the equation, following y=....
  • Round the values in the equation to 2 decimal places if necessary.
Answer: Complete the equation for the line: y = .
one-of
type(expression)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Start by using the inclination to find the gradient of the line. Then use that gradient to start working out the equation.


STEP: Use the inclination to find the gradient of the line
[−1 point ⇒ 3 / 4 points left]

This question is about finding the equation of a straight line. We know one point on the line and the line's inclination. The equation of the line will have the form y=mx+c. So we need to find the gradient, m, and the y-intercept, c.

The inclination of a line tells us about the steepness of the line so it is similar to the gradient. In fact, inclination and gradient are directly related to each other. We can use the following equation to calculate the gradient from the inclination.

m=tanθ=tan(116,565°)=2,00000...2

The gradient is approximately 2 (remember that the instructions state the we should round the values to 2 decimal places). Note that the gradient is negative, which agrees with the slope of the graph.


STEP: Put the gradient into the equation
[−1 point ⇒ 2 / 4 points left]

Now we recall the general form of the equation of a straight line. Since we know the gradient m of the line, we can substitute that into the equation of the line:

y=(2)x+c

STEP: Use the point to find the y-intercept
[−1 point ⇒ 1 / 4 points left]

Now we can calculate the value of the y-intercept of the line (the value of c). We do this by substituting the coordinates of any point on the line into the general form for the line from above. Since we know the coordinates of point A, we will use those coordinates, which are A(3;112).

y=2x+c(112)=2(3)+c0,5=c

Remember that this number should be the y-intercept of the graph. As with the gradient, we can compare this answer to the graph to check if they agree: does the graph up above have a y-intercept close to 0,5? Yes!


STEP: Write the complete equation
[−1 point ⇒ 0 / 4 points left]

We now have the gradient and the y-intercept of the line. The equation of the line is:

y=2x0,5

Submit your answer as:

6. Practical application

Exercises